Algebraic Topology <author>Andries Brouwer, <tt/ <date>v1.0, 991111 <abstract> Some fragments of algebraic topology. </abstract> <sect>Introduction<p> Topology studies spaces up to homeomorphism. Algebraic topology invents coarser invariants, like homotopy type and homology group. These will help distinguish spaces that are sufficiently different, and allows the use of algebraic machinery. Distinguishing spaces is the same as showing that there is no homeomorphism between them. More generally, these algebraic techniques enable one to show that maps with certain properties do not exist (because they would lead to morphisms of homotopy or (co)homology groups, and by algebra these do not exist). <label id="brouwer"> A famous example is the Brouwer fixed point theorem: <bf>Theorem</bf> Let <it>E(n)</it> be the <it>n</it>-ball in Euclidean <it>n</it>-space. Then every continuous map from <it>E(n)</it> into itself has a fixed point. This is an immediate corollary of <bf>Theorem</bf> Let <it>S(n-1)</it> be the <it>(n-1)</it>-sphere in Euclidean <it>n</it>-space, the boundary of <it>E(n)</it>. If <it>n > 0</it> then the identity map from <it>S(n-1)</it> to itself has no extension to a map from <it>E(n)</it> to <it>S(n-1)</it>. <sect> Topology <p> <sect1>Topological spaces<p> A <it>topological space</it> is a set <it>X</it> together with a collection of subsets <it>OS</it> the members of which are called <it>open</it>, with the property that (i) the union of an arbitrary collection of open sets is open, and (ii) the intersection of a finite collection of open sets is open. (In particular <it>X</it> is open, as is the empty set.) <bf>Examples</bf> (i) Metric spaces: there is a realvalued distance function on <it>X × X</it> satisfying the triangle inequality; an open set is a set containing balls with sufficiently small radius around each of its points. <newline> (ii) Ordered spaces: there is a total order on <it>X</it>; an open set is a set containing for each point <it>y</it> an entire orderinterval <it>(x,z)</it>, where <it>x < y < z</it>. <newline> (iii) Discrete topology: every subset of <it>X</it> is open. <newline> (iv) Indiscrete topology: the only open subsets of <it>X</it> are the empty set and <it>X</it> itself. A <it>map</it> or <it>continuous function</it> from a topological space <it>(X,OX)</it> to a topological space <it>(Y,OY)</it> is a function from <it>X</it> to <it>Y</it> such that the preimage of any member of <it>OY</it> is a member of <it>OX</it>. A <it>homeomorphism</it> is a bijective map of which the inverse is a map, too. A <it>neighbourhood</it> of a set is an open set containing it. A <it>closed</it> set is the complement of an open set. A <it>basis</it> for the open sets is a collection of open sets such that each open set is a union of some subcollection. A <it>subbasis</it> for the open sets is a collection of open sets such that one obtains a basis by taking finite intersections. In order to check that a given function is continuous, it suffices to check that the inverse images of the members of a subbasis for the open sets are open again. The <it>interior</it> of a set <it>A</it> is the largest open set contained in <it>A</it>. The <it>closure</it> of <it>A</it> is the smallest closed set containing <it>A</it>. The <it>boundary</it> of <it>A</it> is the intersection of its closure and the closure of its complement. <bf>Example</bf> Let <it>(X,OX)</it> be the closed unit interval [-1,1] with usual topology and let <it>(Y,OY)</it> be the space obtained from it by identifying the points 0 and 1. Let <it>p</it> be the quotient map. Then the restriction <it>p'</it> of <it>p</it> to the half-open interval [-1,1) is a bijection and is continuous, but its inverse is not continuous, so <it>p'</it> is not a homeomorphism. <bf>Exercise</bf> (i) Let <it>A</it> be a subset of a topological space <it>X</it>. How many different subsets can I make starting from <it>A</it> and repeatedly using the closure and interior operations? <htmlurl name="Answer" url="GaloisCorrespondence.html">. <newline> (ii) Let <it>A</it> be a subset of a topological space <it>X</it>. How many different subsets can I make starting from <it>A</it> and repeatedly using the boundary and union operations? <htmlurl name="Answer" url="TopBoundaryUnion.html">. <sect1>Operations on topological spaces<p> <sect2>Topologies defined by functions<p> Let <it>A</it> be a set, and <it>F</it> a collection of functions from <it>A</it> to topological spaces. The topology on <it>A</it> defined by <it>F</it> is the weakest topology (i.e., the smallest collection <it>OA</it>) for which all these functions become continuous. Similarly, let <it>B</it> be a set, and <it>F</it> a collection of functions from topological spaces to <it>B</it>. The topology on <it>B</it> defined by <it>F</it> is the strongest topology (i.e., the largest collection <it>OB</it>) for which all these functions become continuous. <sect2>Subspaces<p> Given a topological space <it>(X,OX)</it> and a subset <it>A</it>, we may (and will) consider it a topological space in its own right (a <it>subspace</it>) by giving it the topology defined by the inclusion map. <sect2>Product spaces<p> Given an arbitrary collection of topological spaces <it>(Xi,OXi)</it>, their (Cartesian) product is the topological space with as point set the Cartesian product of the sets <it>Xi</it>, and topology defined by the projection maps. <sect2>Quotient spaces<p> Given a topological space <it>(X,OX)</it> and a function <it>f</it> from <it>X</it> to a set <it>B</it>, we call the topology on <it>B</it> determined by <it>f</it> the <it>quotient topology</it>, and <it>f</it> the corresponding <it>quotient map</it>. Frequently <it>B</it> will be the set of equivalence classes in <it>X</it> of some equivalence relation <it>R</it>. In this case the quotient space is denoted <it>X/R</it>. <sect1>Separation<p> A topological space is called <it>Hausdorff</it> (or (T2)) when any two points have disjoint neighbourhoods. (A weaker requirement called (T1) is that every singleton is a closed subset.) A topological space is called <it>normal</it> when any two disjoint closed sets have disjoint neighbourhoods. A topological space is called <it>metric</it> when there is a <it>distance function</it> determining the topology (i.e., open balls for the metric are open sets, and conversely, if a point <it>x</it> lies in an open set <it>U</it> then for some positive <it>e</it> the ball with radius <it>e</it> around <it>x</it> is contained in <it>U</it>. A metric space is normal since <it>{x|d(x,A) < d(x,B)}</it> and <it>{x|d(x,A) > d(x,B)}</it> are disjoint neighbourhoods of the disjoint closed sets <it>A</it> and <it>B</it>. A Hausdorff space <it>X</it> is normal if and only if for each pair of disjoint closed sets <it>A</it> and <it>B</it> there exists a map <it>f</it> from <it>X</it> to the unit interval <it>I</it> that is identically 0 on <it>A</it> and identically 1 on <it>B</it>. (Urysohn) <sect1>Covering<p> A topological space is called <it>compact</it> when every open cover (i.e., covering with open sets) has a finite subcover. A compact subset of a Hausdorff space is closed. A closed subset of a compact space is compact. The continuous image of a compact space is compact again. A topological space is called <it>paracompact</it> when every open cover has a locally finite refinement. <sect1>Convergence<p> A <it>filterbase</it> is a collection of nonempty sets such that the intersection of any two contains a third. We say that a filterbase converges to a point if each neighbourhood of the point contains an element of the filterbase. We say that a filterbase accumulates at a point if each neighbourhood of the point meets an element of the filterbase. A space is Hausdorff iff each convergent filterbase converges to exactly one point. A space is compact iff each filterbase has an accumulation point. A space is compact iff each maximal filterbase converges. <bf>Exercise</bf> Prove: the cartesian product of compact spaces is compact. (Tychonoff) <sect1>Connectedness<p> A topological space is called <it>connected</it> when the only subsets that are both open and closed are the empty set and the entire space. A <it>(connected) component</it> of a topological space is a maximal connected subset. The continuous image of a connected space is connected again. <sect2>Path-connectedness<p> In particular, an image of the closed unit interval [0,1] (sometimes called an <it>arc</it> or a <it>path</it>) is connected. A topological space is called <it>path-connected</it> or <it>arcwise connected</it> when any two of its points can be joined by an arc. (Note: below we shall use the word path for the mapping from [0,1] into some space, rather than for its image.) <sect2>Quasi-components<p> The intersection of all open-and-closed subsets containing a given point <it>x</it> is called the <it>quasi-component</it> of <it>x</it>. The partition of a topological space into quasi-components is coarser than the partition into components. <bf>Exercise</bf> Prove: The closure of a connected set in an arbitrary topological space is again connected. Components and quasicomponents of a topological space are closed. <bf>Exercise</bf> Prove: Let <it>X</it> be connected, and <it>A</it> a connected subset, and <it>Q</it> either open-and-closed, or a component, or a quasicomponent in <it>X\A</it>. Then <it>X\Q</it> is connected. <htmlurl name="Answer" url="Component.html"> <bf>Exercise</bf> (i) Construct a topological space that is connected but not path-connected. (ii) Construct a topological space with a quasicomponent that is not connected. (iii) Show that in a compact space components and quasicomponents coincide. <bf>Warning</bf> There exist strange objects, like connected Hausdorff spaces that become totally disconnected (all components are singletons) when one removes a well-chosen point, or connected metric spaces without nontrivial open connected subspaces. Also, there need not be any non-constant maps from [0,1] to a connected topological space (indeed, there exist countable connected Hausdorff spaces), and in such a case all path-components are singletons. <p> <bf>Problem</bf> Show that there exists a connected subset of the Euclidean plane such that all its path-components are singletons. <sect1>Local properties<p> A topological space is said to be <it>locally P</it> for some property <it>P</it> when for each point <it>x</it> and neighbourhood <it>U</it> of <it>x</it> there is a set <it>A</it> contained in <it>U</it> and containing a neighbourhood of <it>x</it> that has property <it>P</it>. For example, the Euclidean plane is locally compact but not compact. <bf>Exercise</bf> (i) Give an example of a space that is connected but not locally connected. (ii) Show that if <it>X</it> is connected and <it>A</it> a compact nonempty proper subspace, then each component of <it>A</it> meets the boundary of <it>A</it>. <sect1>Function spaces<p> Let <it>C(X,Y)</it> be the set of all maps from <it>X</it> into <it>Y</it>. This set has several natural topologies. First of all, there is the product topology, where this is regarded as a subspace of the product of <it>|X|</it> copies of <it>Y</it>. Here a basic open set constrains the values at finitely many points to an open set. Much finer is the <it>compact-open</it> topology (also called <it>c</it>-topology), where a subbasic open set contains the functions mapping a given compact subset of <it>X</it> into a given open subset of <it>Y</it>. Below we shall assume that <it>C(X,Y)</it> has the compact-open topology. Let <it>Y</it> be locally compact and <it>X,Z</it> Hausdorff. Then the composition map from <it>C(X,Y) × C(Y,Z)</it> to <it>C(X,Z)</it> is continuous. In particular, under these assumptions, the evaluation map from <it>C(Y,Z) × Y</it> to <it>Z</it> is continuous. <sect>Homotopy<p> <sect1>Homotopic maps<p> Let <it>X,Y</it> be two topological spaces, and <it>I</it> the closed unit interval [0,1]. Two maps <it>f,g</it> from <it>X</it> to <it>Y</it> are called <it>homotopic</it> if there exists a map <it>F</it> from <it>X × I</it> to <it>Y</it> such that <it>F(x,0) = f(x)</it> and <it>F(x,1) = g(x)</it> for all <it>x</it>. Here <it>F</it> is called a <it>homotopy</it> (from <it>f</it> to <it>g</it>). Intuitively, the second argument can be viewed as time, and then the homotopy describes a continuous deformation from <it>f</it> to <it>g</it>. A map is called <it>nullhomotopic</it> when it is homotopic to a constant map. A space is called <it>contractible</it> when the identity map from the space to itself is nullhomotopic. For example, each real topological vector space is contractible - a suitable homotopy is given by <it>F(x,t) = (1-t)x</it>. More generally, each convex subset of a real topological vector space is contractible. If <it>Y</it> is contractible, then each map from some topological space <it>X</it> to <it>Y</it> is nullhomotopic. Also, if <it>X</it> is contractible, then each map from <it>X</it> into some topological space <it>Y</it> is nullhomotopic. <bf>Exercises</bf> <p> (i) If <it>Y</it> is not pathconnected, then not all constant maps into <it>Y</it> are homotopic. <p> (ii) Let <it>S(n)</it> be the <it>n</it>-dimensional sphere (points of norm 1 in <it>(n+1)</it>-space). If <it>f,g</it> are two maps from some topological space <it>X</it> into <it>S(n)</it>, and for all <it>x</it> the values <it>f(x)</it> and <it>g(x)</it> are not antipodal, then <it>f</it> and <it>g</it> are homotopic. In particular, a nonsurjective map into <it>S(n)</it> is nullhomotopic. <p> (iii) A map <it>f</it> from <it>S(n)</it> into some topological space <it>Y</it> is nullhomotopic if and only if <it>f</it> has a continuous extension to the <it>(n+1)</it>-ball (points of norm at most 1 in <it>(n+1)</it>-space). <sect2>Relative homotopy<p> Let <it>X,Y</it> be two topological spaces, and <it>A</it> a subspace of <it>X</it>. A homotopy <it>F</it> from <it>X × I</it> to <it>Y</it> is called homotopy <it>relative to A</it> if for each <it>a</it> in <it>A</it> the map <it>F(a,t)</it> is constant (independent of <it>t</it>). <sect1>Homotopy classes<p> Being homotopic is an equivalence relation, so we have equivalence classes. Given two spaces <it>X,Y</it>, and a map <it>f</it> from <it>X</it> to <it>Y</it>, let <it>[f]</it> denote the homotopy class of <it>f</it>, that is, the set of all maps from <it>X</it> to <it>Y</it> homotopic to <it>f</it>. Let <it>[X,Y]</it> denote the set of homotopy classes of maps from <it>X</it> to <it>Y</it>. <p> Under suitable assumptions homotopy classes are precisely the path-components in the space <it>C(X,Y)</it> of continuous functions from <it>X</it> to <it>Y</it>. [E.g., give <it>C(X,Y)</it> the compact-open topology, and assume that <it>X</it> is a k-space, i.e., has the topology defined by the injection maps from its compact subspaces. Then this statement holds.] <p> Again, we can look at relative homotopy, and similar things hold. <p> <bf>Exercise</bf> Let <it>f</it> be a path (a map from the unit interval <it>I</it> into some topological space <it>X</it>). Let <it>p</it> be a continuous map from <it>I</it> to itself fixing 0 and 1 (a parameter transformation). Show that <it>[f] = [fp]</it> (where juxtaposition denotes composition: <it>(fp)(t) = f(p(t))</it>). <sect1>Fundamental group<p> We call two paths <it>f,g</it> (maps from <it>I</it> into some space <it>X</it>) homotopic when they are homotopic relative to <it>{0,1}</it>. Given a path <it>f</it>, the <it>inverse path</it> is the path <it>f'</it> defined by <it>f'(t) = f(1-t)</it>. Given two paths <it>f,g</it>, where the endpoint <it>f(1)</it> of the first is the starting point <it>g(0)</it> of the second, their <it>product</it> is the path <it>h = f#g</it> defined by <it>h(t) = f(2t)</it> for <it>2t < 1</it> and <it>h(t) = g(2t-1)</it> otherwise. <bf>Theorem</bf> Let <it>X</it> be a topological space and fix a point <it>x</it> in it. Let <it>e(x)</it> be the constant map that sends all of <it>I</it> to <it>x</it>. Let <it>P(X;x)</it> be the set of homotopy classes of paths starting and ending in <it>x</it>. Then <it>P(X;x)</it> is a group with respect to the multiplication defined by <it>[f][g] = [f#g]</it>. It has inverses defined by <it>[f]' = [f']</it> and unit element <it>[e(x)]</it>. When <it>X</it> is path-connected, all groups <it>P(X;x)</it> are isomorphic. <bf>Proof</bf> We have to check that the operations are well-defined. They are. The group found here is called the <it>Poincaré group</it> or <it>fundamental group</it> of <it>X</it> (at the point <it>x</it>). <bf>Exercise</bf> (i) Show that the fundamental group of the circle <it>S(1)</it> is <it>Z</it>, the additive group of the integers. (ii) Show that the fundamental group of the sphere <it>S(n)</it> with <it>n > 1</it> is trivial. <sect1>Morphisms<p> <bf>Theorem</bf> Let <it>X,Y</it> be two spaces, and let <it>f</it> be a map from <it>X</it> to <it>Y</it>. We find a homomorphism <it>f*</it> from <it>P(X;x)</it> to <it>P(Y;f(x))</it> by sending a path <it>p</it> to the path <it>fp</it> (defined by <it>fp(t) = f(p(t))</it>). This is a functor from pointed topological spaces with maps to groups with homomorphisms. In particular: <it>(gf)* = g* f*</it> and the identity map on <it>X</it> is sent to the identity homomorphism on <it>P(X;x)</it>. It follows that if <it>f</it> is a homotopy equivalence from <it>X</it> to <it>Y</it> (that is, a map such that there is a map <it>g</it> from <it>Y</it> to <it>X</it> such that both <it>fg</it> and <it>gf</it> are nullhomotopic), then <it>f*</it> is an isomorphism. <sect1>Simple connectedness<p> A topological space <it>X</it> is called <it>simply connected</it> when it is path-connected and its fundamental group <it>P(X)</it> is trivial. <bf>Exercise</bf> A path-connected space <it>X</it> is simply connected if and only if each map from the circle <it>S(1)</it> into <it>X</it> can be extended to a map from the 2-ball (circle plus interior) into <it>X</it>. <sect1>The fundamental group of a product<p> <bf>Theorem</bf> The fundamental group of a product is the product of the fundamental groups of the factors. <bf>Proof</bf> If <it>X</it> is a product of factors <it>Xi</it>, and <it>pi</it> is the projection onto <it>Xi</it>, then <it>pi*</it> maps <it>P(X;x)</it> into <it>P(Xi;xi)</it>, and the homomorphism <it>p</it> with <it>i</it>-th coordinate <it>pi</it> maps <it>P(X;x)</it> into the product of the <it>P(Xi;xi)</it>. Now <it>p</it> is trivially surjective and injective, so is an isomorphism. <bf>Exercise</bf> Show that the product of simply connected spaces is simply connected again. <sect1>The fundamental group of a retract<p> Let <it>X</it> be a topological space. A subspace <it>A</it> of <it>X</it> is called a <it>retract</it> of <it>X</it> if there is a map <it>r</it> from <it>X</it> onto <it>A</it> that extends the identity on <it>A</it>. The subspace <it>A</it> is called a <it>deformation retract</it> of <it>X</it> if there is a map <it>r</it> from <it>X</it> onto <it>A</it> that is homotopic to the identity on <it>X</it> relative to <it>A</it>. <bf>Exercise</bf> Clearly, every deformation retract is a retract. Show that the converse does not hold. Given a topological space <it>X</it> and a subspace <it>A</it>, let <it>i</it> be the inclusion map. Let <it>a</it> be a point of <it>A</it>. Then <it>i*</it> is a homomorphism from <it>P(A;a)</it> into <it>P(X;a)</it>. <bf>Exercise</bf> Show that <it>i*</it> need not be injective. <bf>Theorem</bf> If <it>A</it> is a retract of <it>X</it>, then <it>i*</it> is injective. <bf>Theorem</bf> If <it>A</it> is a deformation retract of <it>X</it>, then <it>i*</it> is an isomorphism. <bf>Exercise</bf> Show that a retract of a simply connected space is again simply connected. <sect1>The fundamental group of a union<p> Let <it>X</it> be a topological space that is the union of two path-connected subspaces <it>A</it> and <it>B</it>, where the intersection of <it>A</it> and <it>B</it> is nonempty and path-connected. Then the fundamental group of <it>X</it> is generated by (the images of) the fundamental groups of <it>A</it> and <it>B</it>. More generally we have: <bf>Theorem</bf> (Van Kampen) Under these same hypotheses, let <it>f</it> and <it>g</it> be homomorphisms from <it>P(A;c)</it> (resp. <it>P(B;c)</it>) into some group <it>G</it> that agree on the intersection <it>C</it> of <it>A</it> and <it>B</it>. (Here <it>c</it> is a point of <it>C</it>.) Then there is a unique homomorphism <it>h</it> from <it>P(X;c)</it> into <it>G</it> that agrees with <it>f</it> and <it>g</it> on <it>A</it> and <it>B</it>, respectively. Perhaps more hypotheses are required? <sect1>The fundamental group of a topological group is Abelian<p> If <it>X</it> is a topological group then we have a multiplication of paths given by <it>(f.g)(t) = f(t).g(t)</it> where <it>.</it> denotes the group operation. <bf>Theorem</bf> The group operation on <it>X</it> induces a group operation on <it>P(X;x)</it> that coincides with the old group operation, and <it>P(X;x)</it> is commutative. <bf>Proof</bf> First of all <it>[f].[g] = [f.g]</it> is well-defined. Next, if <it>e = e(x)</it>, then <it>[f].[g] = [f#e].[e#g] = [f#g] = [f][g]</it> and <it>[f].[g] = [e#f].[g#e] = [g#f] = [g][f]</it>. <sect1>The fundamental group of orthogonal groups<p> The fundamental group of <it>SO(2,R)</it>, the group of orthogonal transformations of determinant 1 of the real plane, is isomorphic to <it>Z</it>, the additive group of the integers. (Indeed, this group is isomorphic to <it>S(1)</it>.) The fundamental group of <it>SO(n,R)</it> with <it>n > 2</it> is isomorphic to <it>Z2</it>, the additive group of the integers mod 2. <sect1>Hopf spaces<p> A <it>Hopf space</it> is a space in which the proof given above for the statement that the fundamental group of a topological group is Abelian still works. Thus, by definition, the fundamental group of a Hopf space is Abelian. What do we need? A multiplication <it>.</it> on the space that preserves homotopy so that it induces a multiplication on homotopy classes, and a unit element <it>e</it> for this multiplication such that <it>[e#f] = [f] = [f#e]</it>. <sect1>Higher homotopy groups<p> Instead of using maps from <it>I</it> into <it>X</it>, with the condition that both endpoints are mapped to a given point, we consider maps from <it>I^n</it> into <it>X</it>, with the condition that the entire boundary (the set of all points that have at least one coordinate 0 or 1) is mapped to a given fixed point <it>x</it>. Now again we have a composition # (acting on the first coordinate only) that induces a group operation on the homotopy classes. The group obtained is called <it>pi_n(X;x)</it>. (Thus, <it>pi_1(X;x) = P(X;x)</it>.) Let <it>L(X;x)</it> be the space of loops in <it>X</it> with base point <it>x</it> (i.e., maps from <it>I</it> to <it>X</it> sending 0 and 1 to <it>x</it>), provided with the compact-open topology. For suitably nice (say, locally compact) <it>X</it> we have <it>pi_2(X;x) = P(L(X;x);e)</it> when <it>e</it> is the constant map that sends <it>I</it> to <it>x</it>, and more generally <it>pi_n(X;x) = P(L_(n-1)(X;x);e)</it> where <it>L_n(X;x) = L(L_(n-1)(X;x);e)</it> for a suitable constant <it>e = e_n</it>. For <it>n > 1</it> the group <it>pi_n(X;x)</it> is Abelian. Indeed, <it>L(X;x)</it> is a Hopf space for the operation #. <sect1>Fiber spaces<p> Let <it>E</it> and <it>B</it> be topological spaces (the <it>total</it> and <it>base</it> space, respectively), and let <it>p : E -> B</it> be a map (the <it>projection</it>). We say that <it>p</it> has the CHP (Covering Homotopy Property) for <it>X</it> if for every map <it>F : X×I -> B</it> and every map <it>f' : X -> E</it> with <it>F(x,0) = pf'(x)</it> (for all <it>x</it>) there is a map <it>F' : X×I -> E</it> with <it>F = pF'</it> and <it>F'(x,0) = f'(x)</it> (for all <it>x</it>). In this case <it>(E,B,p)</it> is also called a <it>fiber space</it> (for <it>X</it>). We say that <it>p</it> has the BP (Bundle Property) if there is a space <it>D</it> such that <it>B</it> has an open cover such that for each member <it>U</it> of this cover there is a homeomorphism <it>f_U : U × D -> p^(-1) (U)</it> with <it>p f_U (u,d) = u</it> (for all <it>u,d</it>). In this case <it>(B,E,p)</it> is also called <it>locally trivial</it>. <bf>Theorem</bf> If <it>p</it> has the Bundle Property, then it has the CHP for every paracompact Hausdorff space <it>X</it>. If moreover <it>B</it> is paracompact, then <it>p</it> has the CHP for every topological space. For a proof, see e.g. Dugundji, <it>Topology</it>, Allyn & Bacon, 1966, Chapter XX. If <it>(S(n),B,p)</it> is a fiber space, and <it>B</it> has more than one point, then the map <it>p</it> is not nullhomotopic. Indeed, otherwise the homotopy could be lifted to one from the identity to a map with image contained in a single fiber, so that the identity on <it>S(n)</it> would be nullhomotopic. But it isn't, for example because <it>S(n)</it> has nontrivial homology. Hopf gave as examples locally trivial fiber structures <it>S(3) -> S(2)</it> and <it>S(7) -> S(4)</it> and <it>S(15)->S(8)</it> constructed from the norm 1 elements in the complex, quaternion or octonion plane, mapped to the corresponding point of the projective line. These maps are algebraically trivial, that is, they induce 0 on the homology and cohomology groups, but homotopically nontrivial. Thus, homotopy is strictly finer than homology. <sect1>Some results for spheres<p> A table with some results for spheres, taken from Sze-Tsen Hu, <it>Homotopy Theory</it>, Academic Press, 1959. Let <it>S(n)</it> be the <it>n</it>-sphere, and consider <it>P(m,n) = pi_m(S(n))</it> for <it>n > 0</it>. If <it>m < n</it> then this group is trivial. If <it>m = n</it> then it is isomorphic to <it>Z</it>, the additive group of the integers. If <it>n = 1</it> and <it>m > n</it>, then it is trivial. We have <it>P(m,2) = P(m,3)</it> for every <it>m > 2</it>. If <it>n</it> is odd and <it>m > n</it>, then <it>P(m,n)</it> is finite. If <it>n > 2</it> and <it>m = n+1</it> or <it>m = n+2</it> then it is cyclic of order 2. <it>P(6,3) = Z(12)</it>, the cyclic group of order 12. <it>P(7,4) = Z + Z(12)</it>. If <it>n > 4</it> and <it>m = n+3</it> then <it>P(m,n) = Z(24)</it>. <it>P(7,3) = Z(2)</it>, <it>P(8,4) = Z(2)+Z(2)</it>, <it>P(9,5) = Z(2)</it>. If <it>n > 5</it> and <it>m = n+4</it> then <it>P(m,n) = 0</it>. <it>P(8,3) = Z(2)</it>, <it>P(9,4) = Z(2)+Z(2)</it>, <it>P(10,5) = Z(2)</it>, <it>P(11,6) = Z</it>. If <it>n > 6</it> and <it>m = n+5</it> then <it>P(m,n) = 0</it>. <it>P(9,3) = Z(3)</it>, <it>P(10,4) = Z(24)+Z(2)</it>. If <it>n > 4</it> and <it>m = n+6</it> then <it>P(m,n) = Z(2)</it>. <it>P(10,3) = P(11,4) = Z(15)</it>, <it>P(12,5) = Z(30)</it>, <it>P(13,6) = Z(60)</it>, <it>P(14,7) = Z(120)</it>, <it>P(15,8) = Z+Z(120)</it>. If <it>n > 8</it> and <it>m = n+7</it> then <it>P(m,n) = Z(240)</it>. <it>P(11,3) = P(12,4) = P(13,5) = Z(2)</it>, <it>P(14,6) = Z(24) + Z(2)</it>, <it>P(15,7) = Z(2)+Z(2)+Z(2)</it>, <it>P(16,8) = Z(2)+Z(2)+Z(2)+Z(2)</it>, <it>P(17,9) = Z(2)+Z(2)+Z(2)</it>. If <it>n > 9</it> and <it>m = n+8</it> then <it>P(m,n) = Z(2)+Z(2)</it>. <sect>Two-dimensional manifolds<p> An <it>n</it>-dimensional manifold is a Hausdorff space that is locally isomorphic to Euclidean <it>n</it>-space <it>E(n)</it>. <sect1>Classification<p> The basic examples are the sphere, let us call it <it>M(0)</it>, the torus, let us call it <it>M(1)</it>, and the real projective plane, let us call it <it>N(1)</it>. Arbitrary examples are constructed by glueing. We use the symbol # again. Given two <it>n</it>-manifolds <it>X</it>, <it>Y</it>, and two homeomorphisms <it>f : B -> X</it> and <it>g : B -> Y</it> of the closed unit sphere into these spaces, let <it>X#Y</it> denote the space obtained from the disjoint union of <it>X</it> and <it>Y</it> by removing the images of the open unit sphere, and identifying images of the boundary points in <it>X</it> with corresponding images of boundary points in <it>Y</it>. Clearly, <it>X#M(0) = X</it>. Let <it>M(g) = M(g-1)#M(1)</it> be the sphere with <it>g</it> handles. Let <it>N(h) = N(h-1)#N(1)</it> be the sphere with <it>h</it> crosscaps. <bf>Theorem</bf> Every compact connected 2-manifold is homeomorphic to one of the surfaces <it>M(g)</it> (<it>g > 0</it>) or <it>N(h)</it> (<it>h > 1</it>). In the proof we shall need the concept <it>triangulation</it> of a surface. This is a covering of the space with finitely many triangles, homeomorphic images of an ordinary plane triangle, where two triangles either are disjoint, or have a point or an edge in common, each edge is in precisely two triangles, and the residue at a point is a polygon. <bf>Proof</bf> First of all, these surfaces are all distinct, since they have distinct fundamental groups, see below. Next, since the surface is compact, there is a finite cover with triangles, and after refining where these overlap we find that every compact connected 2-manifold is triangulable. Finally, every connected triangulable 2-manifold is one of these. Indeed, given a triangulation, we can order the triangles such that each after the first has at least one edge in common with an earlier triangle. By induction we see that the subspace on the union of an initial part of this sequence of triangles is homeomorphic to a plane regular polygon of which some pairs of edges are identified. Thus, the entire manifold is isomorphic to a plane <it>2n</it>-gon of which the edges are identified in pairs. Going around the polygon, we can write down a word in <it>n</it> variables <it>x</it> where each occurs twice, either as <it>x</it> or as <it>x^(-1)</it>, written as <it>x'</it>. For example, the word <it>xyx'y'</it> represents the torus. Now we can juggle these words a bit, nothing that (i) cyclic permutation of the word does not change the manifold, (ii) <it>abxcdx' = aydcy'b</it> (where equality denotes that the corresponding 2-manifolds are homeomorphic, and <it>a,b,c,d</it> are arbitrary words), (iii) <it>abxcdx = aydb'yc'</it>. (Indeed, in both cases, let <it>y</it> be an edge from the point where <it>a</it> meets <it>b</it> to the point where <it>c</it> meets <it>d</it>, cut along <it>y</it> and glue along <it>x</it>.) (iv) <it>axx' = a</it> (when <it>a</it> has length at least 4). <newline> Suppose we let <it>a</it> end in a lot of pairs <it>zz</it> of two equal edges. When there are still more equal edges, we can apply (iii) to get the first element of a pair of two equals directly following <it>a</it>, and then once more (with empty <it>b</it> and <it>d</it>) to get both directly following <it>a</it>, and we can enlarge <it>a</it>. After doing this, start moving commutators <it>[x,y] = xyx'y'</it> to the tail of <it>a</it>. When there is an interlocking pair <it>...x...y...x'...y'...</it>, use (ii) to cyclically rotate the parts between <it>y</it> and <it>y'</it> until <it>x</it> is in front of <it>y</it> and <it>x'</it> directly follows it, and then cyclically rotate the part between <it>x'</it> and <it>x</it> until <it>y'</it> directly follows <it>x'</it>, and we have a commutator <it>xyx'y'</it> substring. After all this, our manifold descriptor consists of pairs <it>xx</it> and commutators <it>xyx'y'</it> and pairs <it>xx'</it>. Finally, if both a square and a commutator occur, we can turn this into 3 squares. Indeed, let us rename <it>y</it> to <it>x</it> again, and take two of <it>b,c,d</it> empty, then (iii) says <it>abxx = axb'x = axxb</it>. Now <it>axx[y,z] = axz'y'xy'z' = axyx'yz'z' = axxyyz'z'</it>. Let <it>Gp(x,y,...; e,f,...)</it> denote the group defined by generators <it>x,y,...</it> and relations <it>e = f = ... = 1</it>, and let <it>[x,y]</it> denote the commutator <it>xyx^(-1)y^(-1)</it>. Then <bf>Theorem</bf> (i) <it>P(M(g)) = Gp(x1,y1,...,xg,yg; [x1,y1], ..., [xg,yg])</it> <newline> (ii) <it>P(N(h)) = Gp(x1,...,xh; (x1)^2, ..., (xg)^2)</it>. A different way of distinguishing these surfaces is by use of the Euler characteristic and orientability. The <it>Euler characteristic</it> <it>chi</it> of a triangulable 2-manifold is the number of vertices minus the number of edges plus the number of triangles in any triangulation of the surface. This is well-defined: the number does not change upon refinement of the triangulation, and two triangulations have a common refinement. We have <it>chi(M(g)) = 2-2g</it> and <it>chi(N(h)) = 2-h</it>. More generally, <it>chi(M#N) = chi(M)+chi(N)-2</it>. A triangulable surface is called <it>orientable</it> if it has a triangulation such that we can orient the boundary of all triangles in a directed 3-cycle in such a way that if two triangles have an edge in common they induce opposite directions on this edge. Orientability is invariant under refinement, so orientability is also an invariant of the surface. A surface <it>M#N</it> is orientable if and only if both <it>M</it> and <it>N</it> are. Now the sphere <it>M(0)</it> and the torus <it>M(1)</it> are orientable, but the real projective plane <it>N(1)</it> is not. In the proof we used compactness. Without it there are other examples. The simplest is the ordinary plane. <sect1>The Poincaré Conjecture<p> In the above we saw that nonhomeomorphic compact 2-manifolds are distinguished by their fundamental group. In particular, when a 2-manifold is homotopy equivalent to the sphere <it>S(2)</it>, then it is homeomorphic to the sphere. It is unknown whether the same holds for 3-manifolds and <it>S(3)</it>. This is the famous Poincaré Conjecture. (At first Poincaré conjectured that a compact 3-manifold with the same homology groups as a 3-sphere must be a 3-sphere, but he found a counterexample himself and changed the conjecture to say that any compact 3-manifold with vanishing fundamental group is the 3-sphere.) The 4-dimensional case was settled by Freedman (<it>The topology of four-dimensional manifolds</it>, J. Diff. Geometry <bf>17</bf> (1982) 357-453). All higher dimensional cases had been done already by Smale, see S. Smale, <it>Generalized Poincaré's conjecture in dimensions greater than 4</it>, Ann. Math. <bf>74</bf> (1961) 391-406, and <it>A survey of some recent developments in differential topology</it>, Bull. Am. Math. Soc. <bf>69</bf> (1963) 131-145. Both Smale and Freedman got a Fields medal. <figure><eps file="absent"><img src="mathematicians/Poincare_3.jpeg"></figure> <htmlurl name="Jules Henri Poincaré" url=""> <figure><eps file="absent"><img src="mathematicians/Smale.jpg"></figure> <htmlurl name="Stephen Smale" url=""> <figure><eps file="absent"><img src="mathematicians/Freedman.gif"></figure> <htmlurl name="Michael Hartley Freedman" url=""> <sect1>2-Manifolds with boundary<p> An <it>n</it>-dimensional manifold with boundary is a Hausdorff space where each point has a neighbourhood isomorphic to an open set in a closed halfspace in Euclidean <it>n</it>-space <it>E(n)</it>. From the above we immediately obtain a classification of all 2-dimensional manifolds with boundary. Indeed, the boundary will be a union of disjoint cycles, and if we glue a disk (a `cap') in the hole bounded by a cycle, we fill the hole. After filling all holes we obtain <it>M(g)</it> or <it>N(h)</it>. Thus, the only additional invariant is the number of holes <it>e</it>. <bf>Exercises</bf> (i) Show that a Möbius band (code <it>xpxq</it>) is a real projective plane with one hole. <newline> Show that <it>M(g,e)</it> (<it>M(g)</it> with <it>e</it> holes) has Euler characteristic <it>2-2g-e</it> and that <it>N(h,e)</it> (<it>N(h)</it> with <it>e</it> holes) has Euler characteristic <it>2-h-e</it>. <sect>Knots, Links, Braids<p> A <it>knot</it> is a simple closed curve (homeomorphic image of <it>S(1)</it>) in Euclidean 3-space <it>E(3)</it>. Two knots are called <it>equivalent</it> when there is an orientation-preserving homeomorphism of <it>E(3)</it> onto itself sending one knot to the other. <sect1>Wild embeddings<p> Schoenflies proved in 1908 that any homeomorphism from a simple closed curve in the plane <it>E(2)</it> onto the unit circle <it>S(1)</it> can be extended to a homeomorphism of the plane onto itself. Similar things do not hold in higher dimensions. <figure><eps file="absent"><img src="wildarc.jpg"></figure> For example, there exist <it>wild embeddings</it> of simple arcs into <it>E(3)</it>: homeomorphic images of the unit interval such that the complement is not simply connected. Thus, one usually restricts knots to be tamely embedded, e.g., as a simple closed polygonal curve, and we'll do so as well. Two more examples of wild embeddings: <figure><eps file="absent"><img src="hornedsphere.jpg"></figure> <bf>Example</bf> (Alexander's horned sphere) A homeomorphic image of the sphere <it>S(2)</it> in <it>E(3)</it> such that the complement of the image is not simply connected. <figure><eps file="absent"><img src="antoine.jpg"></figure> <bf>Example</bf> (Antoine's necklace) A homeomorphic image of the Cantor set (which is compact and totally disconnected) in <it>E(3)</it> such that the complement of the image is not simply connected. (The pictures here were taken from Hocking & Young, Topology, pp. 176-177.) <sect1>Knot diagrams and Reidemeister moves<p> Given a knot in <it>E(3)</it>, we project it from a point in general position into <it>E(2)</it>, so that the resulting curve never passes three times through the same point, and indicate for each crossing whether it is an over- or undercrossing. The resulting diagram suffices to retrieve the knot up to equivalence. Now equivalence between knots can be translated into equivalence between diagrams. This was done by Reidemeister, who showed that two diagrams represent the same knot if and only if one is obtained from the other by a sequence of <it>Reidemeister moves</it>: <figure><eps file="absent"><img src="knots/reidemeister.gif"></figure> in each of the three types of move, we may replace the upper picture by the lower, or vice versa; type I also has a mirror image, type I'. There is a unique knot with a diagram without crossings: the <it>unknot</it>. <sect1>Prime knots and Seifert surfaces<p> There is a composition of knots: roughly: tying one after the other. One takes two oriented knots, places two straight line segments with opposite orientation on top of each other, making them cancel. The resulting knot is uniquely determined by the two knots one started with. A knot is called <it>prime</it> if it is not the sum (defined in this way) of two knots, both different from the unknot. Of course this is precisely the 1-dimensional analogue of the # glueing operation used in the previous section. <figure><eps file="absent"><img src="knots/knotsum.gif"></figure> (This picture was taken from Lickorish, An Introduction to Knot Theory, p. 6.) Note that the unknot is a zero element for this sum operation. We have a unique `prime factorization': each knot is a sum of a finite number of prime knots, and the prime knots that go into a knot are uniquely determined. This can be proved by associating a <it>Seifert surface</it> to the knot. A Seifert surface of a link is a compact oriented surface in S(3) with the given link as boundary. Every link has a Seifert surface: give the link some orientation, take a diagram, and replace each crossing by a non-crossing by making the upper strand turn left. Now the diagram has become a union of circuits and we can pick disjoint discs with these circuits as boundaries. Now for each crossing that we replaced, insert a half-twisted strip joining the discs. This yields an oriented surface with the original link as boundary. Now one can define the <it>genus</it> of a knot as the minimal genus of a Seifert surface <it>F</it>for it. (Where one might define the genus in this case by <it>g = (1 - chi(F))/2</it> where <it>chi(F)</it> is the Euler characteristic. The genus is additive: <it>g(K1 + K2) = g(K1) + g(K2)</it>, and is nonnegative, and is zero only for the unknot. It follows that the sum of two knots can only be the unknot when both knots are the unknot themselves. The Seifert surface of a knot is orientable and has a single boundary component, so <it>g = (1 - chi(F))/2 = h</it> is the number of handles, a nonnegative integer, and zero precisely for the unknot. <sect1>Catalog<p> Pictures of the prime knots with a plane drawing with at most 8 crossings (after G. Burde, `Knoten', Jahrbuch Ueberblicke Mathematik, B.I. Mannheim, 1978, pp. 131-147 which contains drawings of the knots with at most 9 crossings. A table of the knots with at most 10 crossings is found in D. Rolfsen, `Knots and Links', Publish or Perish, 1976). The unknot is omitted and orientation is disregarded. <figure><eps file="absent"><img src="knots/03-001.gif"></figure> <figure><eps file="absent"><img src="knots/04-001.gif"></figure> <figure><eps file="absent"><img src="knots/05-001.gif"></figure> <figure><eps file="absent"><img src="knots/05-002.gif"></figure> <figure><eps file="absent"><img src="knots/06-001.gif"></figure> <figure><eps file="absent"><img src="knots/06-002.gif"></figure> <figure><eps file="absent"><img src="knots/06-003.gif"></figure> <figure><eps file="absent"><img src="knots/07-001.gif"></figure> <figure><eps file="absent"><img src="knots/07-002.gif"></figure> <figure><eps file="absent"><img src="knots/07-003.gif"></figure> <figure><eps file="absent"><img src="knots/07-004.gif"></figure> <figure><eps file="absent"><img src="knots/07-005.gif"></figure> <figure><eps file="absent"><img src="knots/07-006.gif"></figure> <figure><eps file="absent"><img src="knots/07-007.gif"></figure> <figure><eps file="absent"><img src="knots/08-001.gif"></figure> <figure><eps file="absent"><img src="knots/08-002.gif"></figure> <figure><eps file="absent"><img src="knots/08-003.gif"></figure> <figure><eps file="absent"><img src="knots/08-004.gif"></figure> <figure><eps file="absent"><img src="knots/08-005.gif"></figure> <figure><eps file="absent"><img src="knots/08-006.gif"></figure> <figure><eps file="absent"><img src="knots/08-007.gif"></figure> <figure><eps file="absent"><img src="knots/08-008.gif"></figure> <figure><eps file="absent"><img src="knots/08-009.gif"></figure> <figure><eps file="absent"><img src="knots/08-010.gif"></figure> <figure><eps file="absent"><img src="knots/08-011.gif"></figure> <figure><eps file="absent"><img src="knots/08-012.gif"></figure> <figure><eps file="absent"><img src="knots/08-013.gif"></figure> <figure><eps file="absent"><img src="knots/08-014.gif"></figure> <figure><eps file="absent"><img src="knots/08-015.gif"></figure> <figure><eps file="absent"><img src="knots/08-016.gif"></figure> <figure><eps file="absent"><img src="knots/08-017.gif"></figure> <figure><eps file="absent"><img src="knots/08-018.gif"></figure> <figure><eps file="absent"><img src="knots/08-019.gif"></figure> <figure><eps file="absent"><img src="knots/08-020.gif"></figure> <figure><eps file="absent"><img src="knots/08-021.gif"></figure> All except the last three are <it>alternating</it>: they have a diagram where along the knot overcrossings and undercrossings alternate. <sect1>Invariants - the Kauffman bracket - the Jones polynomial<p> One of the objects of knot theory is to distinguish inequivalent knots. To this end many invariants have been invented. In this context the most obvious invariant of a knot <it>K</it> is the fundamental group of its complement <it>E(3) \ K</it>. A complete invariant is the complement itself as a topological space: C. Gordon and J. Luecke, `Knots are Determined by their Complements', J. Amer. Math. Soc. 2, 371-415, 1989. Kauffman defines a Laurent polynomial in a variable <it>A</it> (i.e., a polynomial in <it>A</it> and <it>A^(-1)</it>) for each unoriented link diagram <it>D</it> in <it>S(2)</it> as follows: If <it>D</it> is a simple loop, then <it><D> = 1</it>. If <it>D</it> is the disjoint union of <it>D'</it> and a simple loop, then <it><D> = (-A^2-A^(-2))<D'></it>. Otherwise, pick a crossing in <it>D</it> and make diagrams <it>D'</it> and <it>D''</it> by replacing the crossing by two turns where the upper strand turns left in <it>D'</it> and right in <it>D''</it>. Now define <it><D> = A <D'> + A^(-1)<D''></it>. <figure><eps file="absent"><img src="kauffman.jpg"></figure> Then (1) the Kauffman bracket is well-defined, i.e., does not depend on the order in which crossings are eliminated, and (2) the Kauffman bracket is invariant for Reidemeister moves of types II and III. It is not invariant for Reidemeister moves of type I. Indeed, we find <figure><eps file="absent"><img src="kauffman1.jpg"></figure> Now given an oriented link we can do a bit more, and distinguish two types of crossing: those where the top comes from the left (+1) and those where the top comes from the right (-1). Adding the values for all crossings we get the <it>writhe</it> of the diagram. For an unoriented knot this is defined too: the sign of a crossing does not change when the orientation is reversed. <figure><eps file="absent"><img src="writhe.jpg"></figure> The Jones polynomial <it>V(L)</it> of a link <it>L</it> is defined by <it>V(L) = ((-A)^(-3w) <D>)</it> where <it>D</it> is any oriented diagram of <it>L</it> and <it>w</it> is the writhe of <it>D</it>. The usual variable in <it>V(L)</it> is <it>t</it>, where <it>t = A^(-4)</it>. Now <it>V(L)</it> is a polynomial in <it>t</it> when the number of components of <it>L</it> is odd, and in particular when <it>L</it> is a knot. <sect1>Links<p> A <it>link</it> is a disjoint union of simple closed curves in <it>E(3)</it>. Thus, each connected component is a knot, and these knots may be entangled. The smallest few examples are shown here. <p> <figure><eps file="absent"><img src="knots/00-02-01.gif"></figure> <figure><eps file="absent"><img src="knots/02-02-01.gif"></figure> <figure><eps file="absent"><img src="knots/04-02-01.gif"></figure> <figure><eps file="absent"><img src="knots/05-02-01.gif"></figure> <figure><eps file="absent"><img src="knots/06-02-01.gif"></figure> <figure><eps file="absent"><img src="knots/06-02-02.gif"></figure> <figure><eps file="absent"><img src="knots/06-02-03.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-01.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-02.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-03.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-04.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-05.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-06.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-07.gif"></figure> <figure><eps file="absent"><img src="knots/07-02-08.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-01.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-02.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-03.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-04.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-05.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-06.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-07.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-08.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-09.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-10.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-11.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-12.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-13.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-14.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-15.gif"></figure> <figure><eps file="absent"><img src="knots/08-02-16.gif"></figure> (Two links can be nonequivalent but have homeomorphic complements, see C.C. Adams, `The Knot Book: An Elementary Introduction to the Mathematical Theory of Knots', W. H. Freeman, New York, pp. 280-286, 1994.) A link is trivial if and only if the fundamental group of the complement is free. <sect1>Braids<p> A <it>braid</it> on <it>n</it> strings is a collection of <it>n</it> arcs in <it>E(3)</it> starting at the <it>n</it> points <it>(0,j,1)</it> and ending at the <it>n</it> points <it>(0,j,0)</it>, <it>j = 1,2,...,n</it>, each meeting the planes <it>z = c</it> (with <it>0 < c < 1</it>) in a unique point. The elementary braid <it>s(i)</it> is the braid interchanging the two strands numbered <it>i</it> and <it>i+1</it>, where the former overcrosses the latter, where all other strands go straight from top to bottom. We have the obvious concept of composition of braids (similar to that of composition of paths), and find that every braid is (equivalent to) the product of elementary braids. The <it>braid group</it> <it>B(n)</it> of braids with <it>n</it> strings is isomorphic to the group with generators <it>s(i)</it> (<it>i=1,...,n-1</it>) and relations (i) <it>s(i) s(j) = s(j) s(i)</it> when <it>|i-j| > 1</it>, and (ii) <it>s(i) s(i+1) s(i) = s(i+1) s(i) s(i+1)</it> for all <it>i</it>. (Artin) It is possible to view <it>B(n)</it> as the fundamental group of a topological space. (Fox) A theorem by Alexander (J.W. Alexander, `A lemma on systems of knotted curves', Proc. Nat. Acad. Sci. USA 9 (1923) 93-95) shows that every link is equivalent to one obtained from a braid by identifying starting and ending points. <sect>Homology<p> <sect1>Homology of chain complexes<p> A <it>graded Abelian group</it> <it>C = {C_i}</it> is a collection of Abelian groups, indexed by the integers. A <it>homomorphism of degree</it> <it>e</it> from <it>{C_i}</it> to <it>{D_i}</it> is a collection <it>f = {f_i}</it> of homomorphisms of Abelian groups, where <it>f_i : C_i -> D_(i+e)</it>. A <it>morphism</it> of graded Abelian groups is a homomorphism of degree 0. A <it>differential</it> on a graded Abelian group is a homomorphism of some degree with square 0. A <it>chain complex</it> <it>(C,d)</it> is a graded Abelian group <it>C</it> with fixed differential <it>d</it> of degree -1. A <it>morphism</it> of chain complexes (a <it>chain map</it>) is a morphism of graded Abelian groups that commutes with the differential. The <it>homology group</it> <it>H(C)</it> of a chain complex <it>(C,d)</it> is the graded Abelian group <it>H(C) = {H_i(C)}</it> with <it>H_i(C) = ker d_i / im d_(i+1)</it>. (We put <it>Z_i(C) = ker d_i</it>, the <it>cycles</it>, and <it>B_i(C) = im d_(i+1)</it>, the <it>boundaries</it>.) Given a morphism <it>f : C -> D</it> of chain complexes we find an induced morphism <it>f* : H(C) -> H(D)</it>. Thus, <it>H</it> is a functor (with <it>H(f) = f*</it>). <bf>Exercise</bf> Check that <it>f*</it> is well-defined and that <it>H</it> commutes with composition. One has arbitrary sums, products, direct and inverse limits for chain complexes. Taking homology commutes with sums, products and direct limits. <bf>Exercise</bf> Show the homology of an inverse limit may differ from the inverse limit of the homology of a sequence of chain complexes. <sect1>Homology of exterior algebra<p> Given a set <it>V</it>, let the <it>exterior algebra</it> <it>E(V)</it> be the Abelian group generated by associative expressions <it>u_1 ^ ... ^ u_m</it> (where the empty product is allowed and denoted by 1), where the <it>u_i</it> are in <it>V</it>, subject to the relations <it>u ^ u = 0</it> and <it>u ^ v + v ^ u = 0</it>. Let <it>E'(V)</it> be the subgroup of the elements with an expression not involving 1. The group <it>E(V)</it> has a natural grading (say, with <it>u_0 ^ ... ^ u_m</it> having degree <it>m</it>) and a natural differential <it>d</it> of degree -1 defined by <it>d_m(u_0 ^ ... ^ u_m) = Sum_i (-1)^i u_0 ^ ... ^ u_(i-1) ^ u_(i+1) ... ^ u_m</it>. Also <it>E'(V)</it> has a natural differential, defined by the above expression for <it>m > 0</it> and zero on degree 0 and smaller. Thus, we find homology in both cases. <bf>Exercise</bf> Show that <it>E(V)</it> has zero homology. Show that <it>C = E'(V)</it> has all homology groups zero, except for <it>H_0(C)</it>, which is isomorphic to the integers when <it>V</it> is nonempty. [Maybe it feels as if the grading is wrong, shifted by one. I agree. But this is tradition. Similarly, one usually defines homology <it>H</it> using <it>E'(V)</it>, notices that there are formal complications in degree 0, and then looks at the <it>augmented</it> chain complex <it>E(V)</it> with <it>reduced</it> homology <it>H</it>-tilde, which equals <it>H</it> everywhere, except in degree 0, in case <it>V</it> is nonempty, where <it>H_0</it> has an additional summand <it>Z</it>. The theory is simplified by never mentioning <it>E'(V)</it> and only using reduced homology.] <sect1>Simplicial homology<p> A <it>simplicial complex</it> is a set <it>V</it> together with a collection <it>K</it> of nonempty finite subsets called <it>simplices</it> such that any nonempty subset of a simplex (called <it>face</it> of the simplex) is again in <it>K</it>. Each simplicial complex (with underlying set <it>V</it>) gives rise to the subgroup <it>E'(K)</it> of <it>E'(V)</it> generated by the products <it>u_1 ^ ... ^ u_m</it> for each simplex <it>{ u_1, ..., u_m }</it> in <it>K</it>, and <it>E'(K)</it> is invariant for the differential <it>d</it>. The corresponding homology is called simplicial homology. <sect1>Singular homology<p> Let <it>X</it> be a topological space. A <it>singular simplex</it> in <it>X</it> is a map from a Euclidean (solid) simplex (with ordered set of vertices) into <it>X</it>, where two singular simplices are considered equal when they differ by the unique affine transformation of the domain simplices that preserves vertex order. (One does not need a Euclidean space to start with; given an ordered <it>m</it>-set <it>M</it>, one constructs a solid <it>(m-1)</it>-simplex by taking all functions from <it>M</it> to the unit interval with values summing to 1. The coordinates obtained in this way are called <it>barycentric coordinates</it>. Now two singular simplices are considered identical when they only differ by an order preserving bijection of the domains of their domains. Thus, there is no loss in generality to take the set of integers <it>0..m</it> as domain for all <it>m</it>-simplices.) We obtain a simplicial complex (called <it>S(X)</it>) by letting the faces of a singular simplex be the restrictions of the map to the faces of the domain simplex. The homology obtained in this way is called the <it>singular homology</it> of the topological space. If <it>X</it> and <it>Y</it> are topological spaces, and <it>f</it> is a map from <it>X</it> to <it>Y</it>, then composition with <it>f</it> defines a map from <it>S(X)</it> into <it>S(Y)</it> that commutes with taking faces, and hence defines a homomorphism <it>f*</it> (of degree 0) from <it>H(X)</it> to <it>H(Y)</it>. <sect1>.Cech homology<p> [<figure><eps file="absent"><img src="cech.gif"></figure> - unfortunately ISO-8859-1 does not have a hacek.] An entirely different way of finding a simplicial complex in a topological space is by taking an open cover and letting a simplex be a subset of the cover with nonempty intersection. This simplicial complex is called the <it>nerve</it> of the cover. For each cover we obtain a homology group. Given a cover and a refinement, we can map the simplices belonging to the refinement to simplices belonging to the first cover by picking for each open set in the refinement a member of the first cover containing it. This leads to a homomorphism of homology groups. The collection of all open covers is directed for refinement, and we get an inverse system. The inverse limit is the .Cech homology group. (Details later, when we have some more machinery.) <sect1>Chain homotopy<p> Let <it>f,g : C -> D</it> be morphisms of chain complexes. A <it>chain homotopy</it> from <it>f</it> to <it>g</it> is a homomorphism <it>e : C -> D</it> of degree 1 such that <it>f - g = de + ed</it>. If <it>f,g</it> are homotopic, i.e., if there is a chain homotopy from <it>f</it> to <it>g</it>, then <it>f* = g*</it>, i.e., <it>f</it> and <it>g</it> induce the same map <it>f* : H(C) -> H(D)</it>. [Indeed, if <it>z</it> is a cycle in <it>C</it>, then <it>ed(z) = 0</it> and hence <it>f(z) - g(z) = de(z)</it> is a boundary.] <bf>Exercise</bf> Being homotopic is an equivalence relation. If <it>f,g : C -> D</it> are homotopic, and <it>f',g' : D -> E</it> are homotopic, then <it>f' f</it> and <it>g' g</it> are homotopic. A chain complex <it>C</it> is called <it>contractible</it> when the identity on <it>C</it> is homotopic to 0. It follows that <it>H(C) = 0</it>, that is, <it>C</it> is <it>acyclic</it>. For example, a simplex, or, more generally, a simplicial complex all of whose simplices contain some fixed vertex <it>v</it>, is acyclic: the map <it>e</it> defined by <it>e(v_0 ^ ... ^ v_m) = v ^ v_0 ^ ... ^ v_m</it> is a chain homotopy from the identity to 0. Let us look at .Cech homology again. Given an open cover <it>U</it> and a refinement <it>V</it>, there are in general many ways of assigning to each member of <it>V</it> a member of <it>U</it> containing it. Thus, we find many morphisms of the corresponding chain complexes. However, these are all homotopic, and hence induce the same mapping on the homology groups. [Indeed, if the two maps assign <it>u</it> and <it>u'</it> (respectively) to some <it>v</it> in <it>V</it>, then we can define a chain homotopy <it>e</it> by <it>e(v_0 ^ ... ^ v_m) = Sum_i (-1)^i u_0 ^ ... ^ u_i ^ u'_i ^ ... ^ u'_m</it>, so that <it>(de+ed)(v_0 ^ ... ^ v_m) = u'_0 ^ ... ^ u'_m - u_0 ^ ... ^ u_m</it>.] <bf>Exercise</bf> Show that given topological spaces <it>X,Y</it> and a map <it>f:X->Y</it> we find a map <it>f(V)*: H_U(X) -> H_V(Y)</it> on the .Cech homology groups corresponding to open covers <it>U</it> of <it>X</it> and <it>V</it> of <it>Y</it>, where <it>U</it> is the collection of inverse images of <it>V</it>. Show that this induces a map <it>f*: H(X) -> H(Y)</it>, on the .Cech homology groups of <it>X</it> and <it>Y</it> (the inverse limit of the groups <it>H_U(X)</it> and <it>H_V(Y)</it>). Let <it>f : C -> C'</it> be a morphism of chain complexes. The <it>mapping cone</it> of <it>f</it> is the chain complex <it>C''</it> with <it>C''_i = C_(i-1) + C'_i</it> (direct sum) and <it>d'' (c,c') = (-dc, fc+d'c')</it>. Then <it>f</it> is a chain equivalence (i.e., there exists a morphism of chain complexes <it>f' : C' -> C</it> such that both <it>ff'</it> and <it>f'f</it> are homotopic to 1), if and only if <it>C''</it> is contractible. <bf>Exercise</bf> Verify all details. More precisely: (i) <it>C''</it> is a chain complex, and (ii) If <it>C''</it> is contractible, so that <it>d''e''+e''d'' = 1</it> for some <it>e''</it>, then <it>e''(c,0) = (ec,*)</it> and <it>e''(0,c') = (f'c', -e'c')</it> define a morphism of chain complexes <it>f'</it> and chain homotopies <it>e</it> and <it>e'</it> from <it>f'f</it> and <it>ff'</it> to 1. (iii) If <it>f' : C' -> C</it> is a morphism of chain complexes and <it>e</it> and <it>e'</it> are chain homotopies from <it>f'f</it> and <it>ff'</it> to 1, then <it>e''(c,c') = ((e+f'e'f-f'fe)c+f'c', (e'fe-e'e'f)c-e'c')</it> defines a contraction of <it>C''</it>. Let us look at singular homology again. Instead of oriented simplices we can use ordered simplices (where the generating elements are sequences of points, possibly with repetitions, that occur in a common simplex). <bf>Exercise</bf> The map that assigns to an ordered simplex <it>(u_0,...,u_m)</it> the oriented simplex <it>u_0 ^ ... ^ u_m</it> is a chain equivalence. <htmlurl name="Answer" url="ordsimpl.gif"> <bf>Exercise</bf> A simplicial complex <it>K</it> and its barycentric subdivision (that has as vertices the simplices of <it>K</it>, and as <it>m</it>-simplices the chains consisting of <it>i</it>-simplices of <it>K</it> for <it>i=0,...,m</it>, totally ordered by inclusion) are chain equivalent. <sect1>Singular Homology and Homotopy<p> Homotopy is a finer invariant than homology. We give two results in this direction. <bf>Theorem</bf> Let <it>X,Y</it> be two topological spaces, and <it>f,g : X -> Y</it> two homotopic maps. Then we find for singular homology that the maps <it>f*,g* : H(X) -> H(Y)</it> coincide. Let us set up a little bit of machinery, so that the proof will be obvious. Given a solid simplex <it>A</it>, let <it>i,j : A -> A × I</it> be the two injections sending <it>a</it> to <it>(a,0)</it> and <it>(a,1)</it>, respectively. Let us define a canonical subdivision (triangulation) <it>T(A)</it> of <it>A × I</it> as follows. Let <it>c</it> be the center of <it>A</it>. Then <it>T(A)</it> consists of the simplices <it><c,iA></it> and <it><c,jA></it> (and <it><c,iF></it> and <it><c,jF></it> for all faces <it>F</it> of <it>A</it>) and all simplices <it><c,S></it> with <it>S</it> in <it>T(F)</it> for some face <it>F</it> of <it>A</it>. (Given a point <it>x</it> distinct from <it>c</it> in <it>A × I</it>, the line <it>cx</it> hits the boundary of <it>A × I</it> either in <it>iA</it> or in <it>jA</it> or in <it>F × I</it> for some face <it>F</it> of <it>A</it>, so we do indeed cover <it>A × I</it>.) Given a simplicial complex <it>K</it>, let <it>T(K)</it> be the simplicial complex that is the union of <it>T(A)</it> for all simplices <it>A</it> of <it>K</it>. The maps <it>i</it> and <it>j</it> extend linearly to chain maps (denoted by the same symbols) from the chain complex defined by <it>K</it> to that defined by <it>T(K)</it>. <bf>Lemma</bf> The chain maps <it>i,j : K -> T(K)</it> are chain homotopic. <bf>Proof</bf> Consider the chain map <it>k : K -> T(K)</it> that assigns to a simplex <it>A</it> the element <it><c,iA> - <c,jA> - <c,kdA></it> (defined by induction of dim <it>A</it>; <it><c,*></it> extended linearly). We see by induction that <it>dk+kd = i-j</it>. Indeed, <it>(dk+kd)(A) = iA - jA - <c,idA> + <c,jdA> + <c,dkdA> = iA - jA - <c,kddA> = iA - jA</it>. QED <bf>Proof of the theorem</bf> Let <it>i,j : X -> X × I</it> be the two injections sending <it>x</it> to <it>(x,0)</it> and <it>(x,1)</it>, respectively. Since <it>f</it> and <it>g</it> are homotopic, there is a map <it>F : X × I -> Y</it> such that <it>f = Fi</it> and <it>g = Fj</it>. Thus, it suffices to show that <it>i* = j*</it>. We do this by exhibiting a chain homotopy <it>h</it> from <it>i</it> to <it>j</it> considered as maps from <it>S(X)</it> to <it>S(X × I)</it>. Thus, <it>h</it> will be a map from <it>S(X)</it> to <it>S(X × I)</it> of degree 1, such that <it>dh+hd = i-j</it>. Elements of <it>S(X)</it> are (linear combinations of) maps <it>s</it> from a (solid) simplex <it>A</it> into <it>X</it>. Define <it>h(s) = (s × 1) k(A)</it> where <it>k</it> is the map defined in the proof of the lemma, viewed as a linear combination of maps into <it>A × I</it> and 1 is the identity on <it>I</it>. We find <it>(dh+hd)(s) = (s × 1)(dk+kd)(A) = (s × 1)(i-j)(A) = (i-j)(s)</it>. QED The second result here is that H_1 equals pi_1 made Abelian, that is, the first homology group of a space (for singular homology) equals the quotient of the fundamental group by its commutator subgroup. <bf>Theorem</bf> <it>H_1(X)</it> is the quotient of <it>P(X,x)</it> by its commutator subgroup. <bf>Proof</bf> Let <it>P(X;x)</it> be the fundamental group of <it>X</it> and let <it>H_1(X)</it> be the first homology group. We find a homomorphism <it>h : P(X;x) -> H_1(X)</it> as follows. A map <it>f : I -> X</it> with <it>f(0) = f(1) = x</it> is a singular simplex of dimension 1, so we can let <it>h</it> map the homotopy class of <it>f</it> to the homology class of <it>f</it>. This is well-defined: we can view <it>f</it> as a map from the circle <it>S(1)</it> into <it>X</it>, and then find a map <it>f* : H(S(1)) -> H(X)</it>, and as we just saw, <it>f*</it> only depends on the homotopy class of <it>f</it>. Let <it>e : I -> S(1)</it> be the map sending <it>t</it> to <it>exp(2 pi i t)</it>. It is a singular simplex and hence represents an element of <it>H(S(1))</it>. The image <it>f*(e)</it> is an element of <it>H(X)</it>, that has representative <it>f</it>. Thus <it>h</it> is well-defined. Moreover, <it>h</it> is a homomorphism. Indeed, we have to check that <it>h(f#g) = h(f)+h(g)</it>, that is, that <it>f + g - f#g = 0</it> in <it>H_1(X)</it>, that is, that <it>f + g - f#g</it> is a boundary, and clearly it is. Remains to show that the commutator subgroup of <it>P(X,x)</it> is the full kernel of <it>h</it>. To this end we construct a left inverse <it>h'</it> to <it>h</it> from <it>H_1(X)</it> into <it>AP(X,x)</it>, the quotient of <it>P(X,x)</it> by its commutator subgroup. The elements of <it>H_1(X)</it> are equivalence classes of 1-cycles in <it>S(X)</it>, and these equivalence classes of 1-cycles have representatives that have boundary contained in <it>{x}</it>. Given such a 1-cycle <it>z</it>, let <it>h'</it> map it to <it>z</it>, regarded as a path in <it>X</it>. We have to show that this is well-defined, that is, that <it>z</it> is null-homotopic (0 in <it>AP(X,x)</it>) when it is a boundary. But that is clear. QED <sect1>Exact sequences<p> Let <it>A -> B -> C</it> be a chain of Abelian groups, connected by mappings <it>f : A -> B</it> and <it>g : B -> C</it>. The chain is called <it>exact</it> at <it>B</it> when the image of <it>f</it> equals the kernel of <it>g</it>. A longer chain is called exact when it is exact at the middle of each 3-term subsequence. An exact sequence of chain complexes is a sequence of morphisms of chain complexes giving an exact sequence of Abelian groups for each index. <bf>Theorem</bf> Given a short exact sequence <it>0 -> C -> D -> E -> 0</it> of chain complexes, we find a long exact sequence <it> ... -> H_i(C) -> H_i(D) -> H_i(E) -> H_(i-1)(C) -> ...</it>. The correspondence is functorial: a morphism between two short exact sequences is mapped to a morphism between two long exact sequences, where a morphism of exact sequences is a collection of mappings making all squares commute. <bf>Proof</bf> We already saw that a morphism <it>C -> D</it> gives a morphism <it>H(C) -> H(D)</it>. Our first task is to define the map <it>H_i(E) -> H_(i-1)(C)</it>. Given <it>z</it> in <it>Z_i(E)</it>, we find <it>y</it> in <it>C_i(D)</it> of which it is the image. Since <it>dz = 0</it> the element <it>dy</it> has zero image in <it>E_(i-1)</it> and hence is the image of a unique element <it>x</it> in <it>C_(i-1)</it>. Moreover, <it>dx</it> has image <it>ddy = 0</it>, so <it>dx = 0</it>, i.e., <it>x</it> is a cycle. Map the homology class of <it>z</it> to that of <it>x</it>. This is well-defined: if we replace <it>y</it> by <it>y'</it> with the same image <it>z</it>, then <it>y'-y</it> is the image of a <it>x'</it> in <it>C_i</it> and <it>x</it> becomes <it>x+dx'</it>, which lies in the same homology class. And if we change <it>z</it> by a boundary: <it>z' = z + dz''</it> with <it>z''</it> in <it>E_(i+1)</it>, then <it>y</it> is changed by an element <it>dy''</it>, where <it>z''</it> is the image of <it>y''</it>, so that <it>dy</it> is not changed at all. Thus we find a function from <it>H_i(E)</it> to <it>H_(i-1)(C)</it>, and one quickly verifies that it is a homomorphism. Remains to verify exactness at each point, and functoriality. Let us leave that as an exercise. <bf>Exercise</bf> Prove that the above long sequence is exact at each point. Prove that the construction behaves well under morphisms. <sect1>The Maier-Vietoris sequence<p> As a special example of the situation of the previous section we have the case of two simplicial complexes <it>B,C</it> of which the union again is a simplicial complex <it>D</it>. Let their intersection be <it>A</it>. Then we have a short exact sequence <it>0 -> A -> B+C -> D -> 0</it> (where <it>+</it> denotes direct sum) if we define the arrows by mapping <it>a</it> to <it>(a,-a)</it> and <it>(b,c)</it> to <it>b+c</it>. Consequently, we find a long exact sequence, called the <it>Maier-Vietoris sequence</it>, <it>... -> H_i(A) -> H_i(B)+H_i(C) -> H_i(D) -> H_(i-1)(A) -> ...</it>. This enables one to compute the homology <it>H_i(D)</it> of a union of two complexes, given the homology of these complexes and that of their intersection. For example, look at the reduced homology of the n-sphere <it>S = S(n)</it>. Let <it>x,y</it> be two points of <it>S</it>. Then <it>S\{x}</it> and <it>S\{y}</it> are homeomorphic to Euclidean n-space <it>E(n)</it> (by `stereographic projection') and hence are contractible and have zero homology. And <it>S\{x,y}</it> is homeomorphic to <it>E(n)</it> minus a point, which is homotopic to <it>S(n-1)</it>. Thus, <it>H_i(S(n)) = H_(i-1)(S(n-1))</it>. At the bottom we have the empty set <it>S(-1)</it> with reduced homology vanishing everywhere except that <it>H_(-1)</it> is isomorphic to the integers. It follows that <it>H_i(S(n)) = 0</it> for all <it>i</it> except that <it>H_n(S(n))</it> is isomorphic to the integers. <bf>Exercise</bf> Prove that <it>E(m)</it> is not homeomorphic to <it>E(n)</it> when <it>m < n</it>. <bf>Exercise</bf> A map <it>f</it> from <it>S(n)</it> to itself has a (mapping) <it>degree</it>, namely the number <it>d</it> such that <it>f*</it> is multiplication by <it>d</it> on <it>H_n(S(n))</it>. Viewing the complex plane together with a point infinity as <it>S(2)</it>, show that the mapping degree of a polynomial of degree <it>d</it> equals <it>d</it>. Conclude that a nonconstant polynomial has a root. <sect1>Relative homology<p> A <it>topological pair</it> is a pair <it>(X,A)</it> where <it>X</it> is a topological space, and <it>A</it> a subspace. A morphism <it>f : (X,A) -> (Y,B)</it> is a map <it>f : X -> Y</it> such that <it>f(A)</it> is contained in <it>B</it>. The singular chain complex <it>S(X,A)</it> is defined as <it>S(X)/S(A)</it> (singular simplices in <it>X</it> modulo those entirely contained within <it>A</it>). The homology is called <it>H(X,A)</it>. All concepts generalize in an obvious way. Regard <it>X</it> as a pair by identifying it with <it>(X,empty)</it>. <bf>Exercise</bf> Prove that if <it>A</it> is a retract of <it>X</it> then <it>H(X) = H(A) + H(X,A)</it>. <bf>Exercise</bf> Prove that <it>S(n-1)</it> is not a retract of <it>E(n)</it>. Conclude that <ref id="brouwer" name="the two theorems"> from the Introduction hold. <sect1>Betti numbers, Euler characteristic, Lefschetz number, Hopf trace formula, Lefschetz fixed point theorem<p> Let <it>A</it> be an <it>R</it>-module. An element <it>a</it> of <it>A</it> is called a <it>torsion element</it> when <it>ra = 0</it> for some nonzero <it>r</it> in <it>R</it>. <bf>Exercise</bf> Give an example to show that the set of torsion elements need not be a submodule. Show that it is a submodule when <it>R</it> is an integral domain (i.e., is commutative, without zero-divisors, with nonzero 1). Assume that <it>R</it> is an integral domain. Let <it>Tor(A)</it> be the torsion submodule of <it>A</it>, that is, the submodule consisting of all torsion elements. The module <it>A</it> is called <it>torsion-free</it> when <it>Tor(A) = 0</it>. The module <it>A' = A / Tor(A)</it> is torsion free. If <it>A</it> is finitely generated, then <it>A'</it> is free, with a finite basis. The cardinality of such a basis is called the <it>rank</it> of <it>A</it>. Let <it>f : A -> A</it> be a morphism. The <it>trace</it> <it>Tr(f)</it> if the trace of <it>f</it> induced on <it>A'</it> (e.g., the sum of the diagonal elements when written on a basis). The above applies (with <it>R</it> the ring of integers) to Abelian groups. The <it>Betti numbers</it> of a simplicial complex are the ranks of the homology groups <it>H_i</it>. Let <it>C</it> be a finitely generated chain complex, and let <it>f : C -> C</it> be a homomorphism of degree 0. The <it>Lefschetz number</it> <it>L(f)</it> is defined as <it>L(f) = sum (-1)^i Tr(f</it>|<it>C_i)</it>. The <it>Euler characteristic</it> <it>chi(C)</it> is the Lefschetz number of the identity map. <bf>Theorem</bf> (Hopf trace formula) Let <it>C</it> be a finitely generated chain complex, and let <it>f : C -> C</it> be a homomorphism of degree 0. Then <it>L(f) = L(f*)</it>. (In particular, the Euler characteristic is the alternating sum of the Betti numbers.) <bf>Exercise</bf> Prove this. (Hint: do the case of the Euler characteristic first.) <bf>Theorem</bf> (Lefschetz fixed point theorem) Let <it>X</it> be a nice compact topological space and let <it>f : X -> X</it> be a map. If <it>L(f)</it> is nonzero then <it>f</it> has a fixed point. (Proof: "nice" here may mean carrier of a finite simplicial complex, or connected orientable <it>n</it>-dimensional manifold, or so. Use a suitable simplicial approximation. If <it>f</it> has no fixed points then all simplices will be moved far away by <it>f</it>, so that <it>f</it> will have trace 0.) <sect1>Axiomatic homology<p> Thee are many homology theories (we have seen singular homology and .Cech homology), and it is possible to develop the theory axiomatically. See S. Eilenberg & N.E. Steenrod, <it>Foundations of Algebraic Topology</it>, Princeton, 1952. A <it>homology theory</it> is a pair <it>(H,d)</it>, where <it>H</it> is a covariant functor from the category of topological pairs to that of graded Abelian groups (i.e., <it>H</it> commutes with composition of maps and sends the identity to the identity), and <it>d</it> assigns to each pair <it>(X,A)</it> a homomorphism of degree -1 from <it>H(X,A)</it> to <it>H(A)</it>, where for <it>f : (X,A) -> (Y,B)</it> we have <it>d(Y,B) H(f) = H(f</it>|<it>A) d(X,A)</it>, satisfying the following four axioms: <newline> <bf>Homotopy Axiom</bf> If <it>f,g : (X,A) -> (Y,B)</it> are homotopic, then <it>H(f) = H(g)</it>. <newline> <bf>Exactness Axiom</bf> Consider the pair <it>(X,A)</it> and the two inclusion maps <it>A -> X -> (X,A)</it>. We find an exact sequence <it>... -> H_i(A) -> H_i(X) -> H_i(X,A) -> H_(i-1)(A) -> ...</it> (with arrows given by <it>...,H,H,d,...</it>). <newline> <bf>Excision Axiom</bf> If <it>U</it> is an open subset of <it>X</it> with closure contained in the interior of <it>A</it>, then the excision map <it>i : (X\U,A\U) -> (X,A)</it> induces an isomorphism <it>H(i)</it> between <it>H(X\U,A\U)</it> and <it>H(X,A)</it>. <newline> <bf>Dimension Axiom</bf> If <it>X</it> is a single point, then <it>H_i(X) = 0</it> for nonzero <it>i</it> and <it>H_0 (X)</it> is isomorphic to the additive group of integers. <sect>Cohomology<p> A <it>cochain complex</it> <it>(C,u)</it> is a graded Abelian group <it>C</it> with fixed differential <it>u</it> of degree 1. (I'll use <it>d</it> for `down' and <it>u</it> for `up'. Standard symbols are the `partial derivative' and `delta' symbols.) We have cocycles and coboundaries and cohomology groups, just as before. A chain complex becomes a cochain complex if we reverse the indexing. Also, the functor <it>Hom(*,G)</it> for fixed <it>G</it> turns chain complexes into cochain complexes and vice versa. (If <it>u</it> is the image of <it>d</it> for this functor, then <it>(us)c = s(dc)</it> for <it>c</it> in <it>C_(i+1)</it> and <it>s</it> in <it>C^i = Hom(C_i,G)</it>. Clearly <it>uu = 0</it>.) Put <it>C* = Hom(C,G)</it> and <it>C^i = (C*)_i</it> and <it>H^i(C) = H_i(C*)</it>. The pairing <it>(s,c) -> sc</it> on <it>C^i × C_i</it> induces a pairing <it>H^i(C) × H_i(C) -> G</it>. [Indeed, if <it>s</it> is a cocycle, that is, <it>us = 0</it>, and <it>c</it> is a boundary, say <it>c = db</it>, then <it>sc = sdb = usb = 0</it>, and if <it>s</it> is a coboundary, say <it>s = ut</it> and <it>c</it> is a cycle, then <it>sc = utc = tdc = 0</it>.] <sect1>The cup product<p> Applying the functor <it>Hom(*,R)</it> where <it>R</it> is a ring, we obtain cohomology with a ring structure derived from that of <it>R</it>. Let us first look at simplicial cohomology over <it>R</it>. Given a <it>i</it>-cochain <it>a</it> and a <it>j</it>-cochain <it>b</it>, we define the <it>(i+j)</it>-cochain <it>a cup b</it> by <it>(a cup b)(v_0 ... v_(i+j)) = a(v_0 ... v_i) b(v_i ... v_(i+j))</it>. Then <it>cup</it> is bilinear, associative, has unit element <it>e</it> in <it>C^0</it> defined by <it>ev = 1</it> for all <it>v</it> in <it>C_0</it>, and satisfies <it>u(a cup b) = ua cup b + (-1)^i a cup ub</it> for <it>a</it> in <it>C^i</it>. This cup product induces a product on the cohomology. [Indeed, if <it>a</it> and <it>b</it> are cocycles, then <it>a cup b</it> is a cocycle, and if moreover <it>a</it> or <it>b</it> is a coboundary, then <it>a cup b</it> is a coboundary.] The resulting structure is called the cohomology ring. If <it>R</it> is commutative, it satisfies the Grassmann property: <it>a cup b = (-1)^(ij) b cup a</it>. On the cochain complex this product was defined for ordered simplices, but in the cohomology ring the product is independent of vertex ordering and we find a cap product also in the cohomology of oriented simplices. One can do the same in general, instead of only for simplicial cohomology, but this requires some more machinery. The idea is that there is a map <it>H*(X) tensor H*(Y) -> H*(X × Y)</it> and composing this (for the case <it>X = Y</it>) with the map <it>H*(X × X) -> H*(X)</it> obtained from the diagonal map <it>X -> X × X</it> we obtain a bilinear multiplication on <it>H*(X)</it>. <sect1>The cap product<p> There is a construct very similar to the cap product which yields a product between homology and cohomology classes. We do the simplicial case again. Given a <it>j</it>-cochain <it>a</it> and an <it>(i+j)</it>-chain <it>v = v_0 ... v_(i+j)</it>, we define the <it>i</it>-chain <it>a cap v</it> by <it>a cap v = a(v_i ... v_(i+j)) v_0 ... v_i</it>. Now <it>cap</it> is bilinear, and associative in the sense that <it>a cap (b cap v) = (a cup b) cap v</it>, has unit element <it>e</it>, and satisfies <it>d(a cap v) = (-1)^i (ua) cap v + a cap dv</it>. This last relation implies that we have a pairing <it>H^j (K;R) × H_(i+j) (K;R) -> H_i (K;R)</it>, and if <it>f</it> is a map then <it>f_*(f^*a cap v) = a cap f_*v</it>. <sect>Products<p> Let all modules be <it>R</it>-modules for some principal ideal domain <it>R</it> and let tensor products be over <it>R</it>. <sect1>The product of two chain complexes<p> Given a product operation like tensor (or *, see below), we can form the product of two chain complexes <it>A</it> and <it>B</it>, say <it>A tensor B</it>, by taking all products <it>A_i tensor B_j</it> and letting <it>(A tensor B)_k = directsum A_i tensor B_j</it> where the sum is over the <it>i,j</it> with <it>i+j=k</it>. We also need a differential, and define <it>d(a tensor b) = da tensor b + (-1)^i a tensor db</it> when <it>a</it> lies in <it>A_i</it>. (Indeed, <it>dd = 0</it>.) There is a natural map <it>H(A) tensor H(B) -> H(A tensor B)</it> sending <it>a tensor b</it> to <it>a tensor b</it>. (Indeed, if <it>a</it> is a boundary, say <it>a = dc</it>, and <it>b</it> is a cycle, then <it>a tensor b = d(c tensor b)</it> is a boundary, so this map is well-defined.) <sect1>The Eilenberg-Zilber theorem<p> For a topological space <it>X</it>, let us write <it>S(X)</it> for the singular chain complex of <it>X</it>. <bf>Theorem</bf> Let <it>X</it> and <it>Y</it> be topological spaces. Then <it>S(X × Y)</it> is chain equivalent to <it>S(X) tensor S(Y)</it>. <sect1>Künneth formula<p> Let <it>C,D</it> be chain complexes, where <it>C*D</it> is acyclic. Then there is a short split exact sequence <it>0 -> H(C) tensor H(D) -> H(C tensor D) -> (H(C) * H(D)) -> 0</it> with maps of degree 0, 0, -1, respectively. This expresses <it>H(C tensor D)</it> in terms of <it>H(C)</it> and <it>H(D)</it>. <sect1>Homology with coefficients<p> Instead of doing homology with integral coefficients, we can use coefficients from an Abelian group <it>G</it>. The resulting chain complexes is <it>C tensor G</it>. The (co)homology obtained in this way is denoted <it>H(C;G)</it>. Let <it>C</it> be a free chain complex (i.e., each <it>C_i</it> is free). Then there is a short split exact sequence <it>0 -> H(C) tensor G -> H(C tensor G) -> H(C) * G -> 0</it> with maps of degree 0, 0, -1, respectively. This expresses <it>H(C;G) = H(C tensor G)</it> in terms of <it>H(C)</it>. <sect1>The cross product<p> There is a homomorphism <it>H(A;G) tensor H(B;H) -> H(A tensor B; G tensor H)</it> defined in the obvious way (using the identification <it>(A tensor G) tensor (B tensor H) = (A tensor B) tensor (G tensor H)</it>). This gives rise to a multiplication on homology called the <it>cross product</it>. Similarly, one has a cross product on cohomology. In particular, following Eilenberg-Zilber, there is for a topological space <it>X</it> a map <it>H*(X;G) tensor H*(X;H) -> H*(X;G tensor H)</it>. But the diagonal map <it>x -> (x,x)</it> from <it>X</it> into <it>X × X</it> gives rise to a homomorphism <it>H*(X × X; K) -> H*(X; K)</it>, and composing that with the cross product we get the <it>cup product</it> <it>H*(X;G) tensor H*(X;H) -> H*(X; G tensor H)</it>, and, in particular, <it>H*(X) tensor H*(X) -> H*(X)</it>. The particular incarnation of the cup product that we described above is obtained by taking the map <it>t : S(X) -> S(X) tensor S(X)</it> defined by <it>t(u) = sum head(u,i) tensor tail(u,j)</it> where the sum is over the <it>i,j</it> with <it>i+j=k</it> and <it>head(u_0...u_k, i) = u_0...u_i</it> and <it>tail(u_0...u_k, j) = u_(k-j)...u_k</it>. <sect>Homological algebra<p> Above we used the undefined product <it>*</it>. It is defined by <it>A*B = Tor_1(A,B)</it>, the <it>torsion product</it> of <it>A</it> and <it>B</it>. It is zero when either <it>A</it> or <it>B</it> is free, and we have <it>A*B = B*A</it>. Just for completeness the definitions. A <it>resolution</it> of <it>A</it> is an exact sequence <it>...-> C_i -> ... -> C_0 -> A -> 0</it>. It is called <it>free</it> when all <it>C_i</it> are free. Free resolutions exist, and any two free resolutions are chain homotopic. Now <it>Tor_i(A,B) = H_i(C tensor B)</it> is independent of the choice of free resolution <it>C</it>. (And dually, <it>Ext^i(A,B) = H^i(Hom(A,B))</it>.) Over a principal ideal domain one finds <it>Tor_i(A,B) = Ext^i(A,B) = 0</it> for <it>i > 1</it>. Also, <it>Tor_0(A,B) = A tensor B</it>, so the only torsion module of interest here is <it>A*B</it>. </article>