MAS61015 Algebraic Topology 13 Chain complexes and homology 15 Homology of spheres

14. Chain homotopy

We next need to build a connection between the concept of homotopy in topology, and the behaviour of chain complexes and homology in algebra.

Video (Definition 14.1 to Proposition 14.7)

Definition 14.1.

Let $A_{*}$ and $A^{\prime}_{*}$ be chain complexes, and let $f,g\colon A_{*}\to A^{\prime}_{*}$ be chain maps (so $df=fd$ and $dg=gd$ ). A chain homotopy between $f$ and $g$ is a system of maps $s_{r}\colon A_{r}\to A^{\prime}_{r+1}$ such that $g_{r}-f_{r}=d^{A^{\prime}}_{r+1}s_{r}+s_{r-1}d^{A}_{r}$ for all $r$ (or more briefly, $g-f=ds+sd$ ). In the case $r=0$ , we should interpret $s_{-1}$ as $0$ , so the condition is $g_{0}-f_{0}=d_{1}s_{0}$ . We say that $f$ and $g$ are chain homotopic if there is a chain homotopy between them. If so, we write $f\cong g$ .

Remark 14.2.

In Section 2 we introduced a crude intuitive version of homology involving chains as subsets of a space $X$ . For such a chain $u\subseteq X$ we can define $\sigma(u)=[0,1]\times u$ , which is a chain in $[0,1]\times X$ ; we call this the thickening of $u$ . If $u$ is a filled triangle in $X$ , then $\sigma(u)$ is a triangular prism. Now $\partial\sigma(u)$ is the boundary of this triangular prism, which consists of the top, the bottom and the sides. On the other hand, $\sigma\partial(u)$ is what we get by thickening the boundary of $u$ , which is just the sides of the prism. After adjusting the $\pm$ -signs to account for orientations, we end up with the relation

\partial\sigma(u)+\sigma\partial(u)=\text{top}-\text{bottom}.

This is a relation between chains in $[0,1]\times X$ , but if we have two maps $f,g\colon X\to Y$ and a homotopy $h\colon[0,1]\times X\to Y$ between them, then we can apply $h_{*}$ to get a relation between chains in $Y$ . The main point of this section is to provide a rigorous and general version of this picture.

Example 14.3.

In Example 13.7 we introduced a chain complex $A$ with

A_{k}=\begin{cases}{\mathbb{Z}}\{e_{i}\;|\;i\in{\mathbb{Z}}/n\}&\text{ if }k=1% \\ {\mathbb{Z}}\{v_{i}\;|\;i\in{\mathbb{Z}}/n\}&\text{ if }k=0\\ 0&\text{ otherwise. }\end{cases}

The differential is given by $d(e_{i})=v_{i+1}-v_{i}$ and $d(v_{i})=0$ . Define $f\colon A_{*}\to A_{*}$ by $f(e_{i})=e_{i+1}$ and $f(v_{i})=v_{i+1}$ . This is a chain map because $d(f(e_{i}))=v_{i+2}-v_{i+1}=f(d(e_{i}))$ . We claim that $f$ is chain homotopic to the identity. Indeed, we can define $s_{0}\colon A_{0}\to A_{1}$ by $s_{0}(v_{i})=e_{i}$ , and we define $s_{i}\colon A_{i}\to A_{i+1}$ to be zero for all $i\neq 0$ . We then find that

	$\displaystyle(ds+sd)(e_{i})$	$\displaystyle=d(0)+s(v_{i+1}-v_{i})=e_{i+1}-e_{i}=f(e_{i})-\operatorname{id}(e% _{i})$
	$\displaystyle(ds+sd)(v_{i})$	$\displaystyle=d(e_{i})+s(0)=v_{i+1}-v_{i}=f(v_{i})-\operatorname{id}(v_{i})$

Proposition 14.4.

The relation of being chain homotopic is an equivalence relation.

Proof.

The zero map is a chain homotopy from $f$ to itself. If $s$ is a chain homotopy from $f$ to $g$ , then $-s$ is a chain homotopy from $g$ to $f$ . If $t$ is also a chain homotopy from $g$ to $h$ , then $s+t$ is a chain homotopy from $f$ to $h$ . ∎

Proposition 14.5.

Suppose we have chain maps

Suppose that $f_{0}$ is chain homotopic to $f_{1}$ , and that $g_{0}$ is homotopic to $g_{1}$ . Then $g_{1}\circ f_{1}$ is chain homotopic to $g_{0}\circ f_{0}$ .

Proof.

We are assuming that $f_{0}$ is chain homotopic to $f_{1}$ , which means that there is a chain homotopy $s$ with $f_{1}-f_{0}=ds+sd$ . Similarly, there is a chain homotopy $t$ with $g_{1}-g_{0}=dt+td$ . Put $u=g_{1}s+tf_{0}$ . Using $dg_{1}=g_{1}d$ we get $du=g_{1}ds+dtf_{0}$ . Using $f_{0}d=df_{0}$ we get $ud=g_{1}sd+tdf_{0}$ . By adding these, we get

du+ud=g_{1}(ds+sd)+(td+dt)f_{0}=g_{1}(f_{1}-f_{0})+(g_{1}-g_{0})f_{0}=g_{1}f_{% 1}-g_{1}f_{0}+g_{1}f_{0}-g_{0}f_{0}=g_{1}f_{1}-g_{0}f_{0},

as required. ∎

Definition 14.6.

We write $\operatorname{hChain}(A_{*},A^{\prime}_{*})$ for the set of chain homotopy classes of chain maps from $A_{*}$ to $A^{\prime}_{*}$ , or in other words equivalence classes under the equivalence relation defined above. Using Proposition 14.5, we see that these are the morphism sets of a well-defined category $\operatorname{hChain}$ , whose objects are chain complexes. This is analogous to the category $\operatorname{hTop}$ introduced in Definition 9.9.

Proposition 14.7.

Let $A_{*}$ and $A^{\prime}_{*}$ be chain complexes, and let $f,g\colon A_{*}\to A^{\prime}_{*}$ be chain maps that are chain homotopic. Then the induced maps $f_{*},g_{*}\colon H_{*}(A)\to H_{*}(A^{\prime})$ are the same.

Proof.

Let $s$ be a chain homotopy from $f$ to $g$ , so $g-f=ds+sd$ . Consider an element $[z]\in H_{r}(A)$ , so $z\in A_{r}$ with $dz=0$ . Recall from Proposition 13.11 that $f(z),g(z)\in Z_{r}(A^{\prime})$ so that the expressions $[f(z)]$ and $[g(z)]$ are meaningful and refer to elements of $H_{r}(A^{\prime})$ . By definition we have $f_{*}[z]=[f(z)]$ and $g_{*}[z]=[g(z)]$ , so we need to check that these are the same. We have $g(z)-f(z)=d(s(z))+s(d(z))$ but $d(z)=0$ so $g(z)=f(z)+d(s(z))\in f(z)+B_{r}(A^{\prime})$ . It follows that $[g(z)]=[f(z)]$ in $H_{r}(A^{\prime})$ , or in other words $f_{*}[z]=g_{*}[z]$ . This means that $f_{*}=g_{*}$ as claimed. ∎

Proposition 14.8.

Let $X$ and $Y$ be topological spaces, and let $f,g\colon X\to Y$ be continuous maps that are homotopic to each other. Then the chain maps $f_{\#},g_{\#}\colon C_{*}(X)\to C_{*}(Y)$ are chain homotopic to each other, so the induced maps $f_{*},g_{*}\colon H_{*}(X)\to H_{*}(Y)$ are the same.

For the proof, we will first choose a homotopy $h$ from $f$ to $g$ , so $h$ is a continuous map $[0,1]\times X\to Y$ with $h(0,x)=f(x)$ and $h(1,x)=g(x)$ for all $x\in X$ . We need to use this to construct a chain homotopy between $f_{\#}$ and $g_{\#}$ , or equivalently a system of maps $\sigma_{k}\colon C_{k}(X)\to C_{k+1}(Y)$ with $\partial\sigma+\sigma\partial=g_{\#}-f_{\#}$ . Before giving the general proof, we will discuss the cases $k=0$ and $k=1$ .

Consider a point $a\in S_{0}(X)=X$ . We can define a continuous map $v\colon\Delta_{1}\to Y$ by $v(1-t,t)=h(t,a)$ . This can be regarded as an element of $S_{1}(Y)\subset C_{1}(Y)$ , and it satisfies

\partial(v)=v(e_{1})-v(e_{0})=h(1,a)-h(0,a)=g(a)-f(a)=g_{\#}(a)-f_{\#}(a).

We define $\sigma_{0}(a)=v$ and extend linearly to get a homomorphism $\sigma_{0}\colon C_{0}(X)\to C_{1}(Y)$ with $\partial(\sigma_{0}(u))=g_{\#}(u)-f_{\#}(u)$ for all $u\in C_{0}(X)$ .

Now consider instead an element $u\in S_{1}(X)$ , or in other words, a continuous map $u\colon\Delta_{1}\to X$ . We want to define $\sigma_{1}(u)\in C_{2}(Y)$ , so $\sigma_{1}(u)$ should be a ${\mathbb{Z}}$ -linear combination of continuous maps from the triangle $\Delta_{2}$ to $Y$ . We have a map $m\colon[0,1]\times\Delta_{1}\to Y$ given by $m(t,s)=h(t,u(s))$ . Here $\Delta_{1}$ is homeomorphic to $[0,1]$ so $[0,1]\times\Delta_{1}$ is a square, with corners $(0,e_{0})$ , $(0,e_{1})$ , $(1,e_{0})$ and $(1,e_{1})$ . We can divide this square into two triangles and restrict $k$ to these triangles, giving two different maps $\Delta_{2}\to Y$ , which will be the terms in $\sigma_{1}(u)$ . In detail, we define maps $\zeta_{0},\zeta_{1}\colon\Delta_{2}\to[0,1]\times\Delta_{1}$ by

	$\displaystyle\zeta_{0}(t_{0},t_{1},t_{2})$	$\displaystyle=t_{0}(0,e_{0})+t_{1}(1,e_{0})+t_{2}(1,e_{1})$
	$\displaystyle\zeta_{1}(t_{0},t_{1},t_{2})$	$\displaystyle=t_{0}(0,e_{0})+t_{1}(0,e_{1})+t_{2}(1,e_{1}).$

The composites $(\Delta_{2}\xrightarrow{\zeta_{i}}[0,1]\times\Delta_{1}\xrightarrow{m}Y)$ (for $i=0,1$ ) can be regarded as elements of $S_{2}(Y)\subset C_{2}(Y)$ . We define $\sigma_{1}(u)=m\zeta_{0}-m\zeta_{1}\in C_{2}(Y)$ . This can be extended linearly to give a homomorphism $\sigma_{1}\colon C_{1}(X)\to C_{2}(Y)$ . We claim that $\partial\sigma_{1}(u)+\sigma_{0}\partial(u)=g_{\#}(u)-f_{\#}(u)$ for all $u\in C_{1}(X)$ . Indeed, it will be enough to prove this when $u\in S_{1}(X)$ . We then have

\partial\sigma_{1}(u)=\partial(m\zeta_{0})-\partial(m\zeta_{1})=m\zeta_{0}% \delta_{0}-m\zeta_{0}\delta_{1}+m\zeta_{0}\delta_{2}-m\zeta_{1}\delta_{0}+m% \zeta_{1}\delta_{1}-m\zeta_{1}\delta_{2}

(where $m(t,s)=h(t,u(s))$ as before). Here, for $t=(t_{0},t_{1})=(1-t_{1},t_{1})\in\Delta_{1}$ we have

	$\displaystyle(\zeta_{0}\delta_{0})(t)$	$\displaystyle=\zeta_{0}(0,t_{0},t_{1})=t_{0}(1,e_{0})+t_{1}(1,e_{1})=(1,t)$
	$\displaystyle(\zeta_{0}\delta_{1})(t)$	$\displaystyle=\zeta_{0}(t_{0},0,t_{1})=t_{0}(0,e_{0})+t_{1}(1,e_{1})=(t_{1},t)$
	$\displaystyle(\zeta_{0}\delta_{2})(t)$	$\displaystyle=\zeta_{0}(t_{0},t_{1},0)=t_{0}(0,e_{0})+t_{1}(1,e_{0})=(t_{1},e_% {0})$
	$\displaystyle(\zeta_{1}\delta_{0})(t)$	$\displaystyle=\zeta_{1}(0,t_{0},t_{1})=t_{0}(0,e_{1})+t_{1}(1,e_{1})=(t_{1},e_% {1})$
	$\displaystyle(\zeta_{1}\delta_{1})(t)$	$\displaystyle=\zeta_{1}(t_{0},0,t_{1})=t_{0}(0,e_{0})+t_{1}(1,e_{1})=(t_{1},t)$
	$\displaystyle(\zeta_{1}\delta_{2})(t)$	$\displaystyle=\zeta_{1}(t_{0},t_{1},0)=t_{0}(0,e_{0})+t_{1}(0,e_{1})=(0,t).$

This can be displayed as follows:

It follows that

	$\displaystyle m\zeta_{0}\delta_{0}(t)$	$\displaystyle=h(1,u(t))=g(u(t))$
	$\displaystyle m\zeta_{0}\delta_{1}(t)$	$\displaystyle=h(t_{1},u(t))=m\zeta_{1}\delta_{1}(t_{0},t_{1})$
	$\displaystyle m\zeta_{0}\delta_{2}(t)$	$\displaystyle=h(t_{1},u(e_{0}))=\sigma_{0}(u(e_{0}))(t)$
	$\displaystyle m\zeta_{1}\delta_{0}(t)$	$\displaystyle=h(t_{1},u(e_{1}))=\sigma_{0}(u(e_{1}))(t)$
	$\displaystyle m\zeta_{1}\delta_{2}(t)$	$\displaystyle=h(0,u(t))=f(u(t)).$

Thus, in our formula for $\partial(\sigma_{1}(u))$ we see that the first and last terms give $g_{\#}(u)-f_{\#}(u)$ , the second and fifth terms cancel out, and the third and fourth terms give $\sigma_{0}(u(e_{0})-u(e_{1}))=-\sigma_{0}(\partial(u))$ . Putting this together, we get $\partial\sigma_{1}(u)+\sigma_{0}\partial(u)=g_{\#}(u)-f_{\#}(u)$ as required.

We now extend the above discussion to cover $k>1$ .

Video (Definition 14.9 to Lemma 14.12)

Definition 14.9.

For $0\leq i\leq k$ we define $\zeta_{i}\colon\Delta_{k+1}\to[0,1]\times\Delta_{k}$ by

	$\displaystyle\zeta_{i}(t_{0},\dotsc,t_{k+1})$	$\displaystyle=\sum_{j=0}^{i}t_{j}.(0,e_{j})+\sum_{j=i+1}^{k+1}t_{j}.(1,e_{j-1})$
		$\displaystyle=t_{0}(0,e_{0})+\dotsb+t_{i}(0,e_{i})+t_{i+1}(1,e_{i})+\dotsb+t_{% k+1}(1,e_{k})$
		$\displaystyle=\left(t_{i+1}+\dotsb+t_{k+1},(t_{0},\dotsc,t_{i-1},t_{i}+t_{i+1}% ,t_{i+2},\dotsc,t_{k+1})\right)$

If it is necessary to specify $k$ , we will write $\zeta_{k,i}$ instead of $\zeta_{i}$ .

Example 14.10.

When $k=2$ the maps $\zeta_{i}\colon\Delta_{3}\to[0,1]\times\Delta_{2}$ are given by

	$\displaystyle\zeta_{0}(x_{0},x_{1},x_{2},x_{3})$	$\displaystyle=(x_{1}+x_{2}+x_{3},(x_{0}+x_{1},x_{2},x_{3}))$
	$\displaystyle\zeta_{1}(x_{0},x_{1},x_{2},x_{3})$	$\displaystyle=(x_{2}+x_{3}+x_{4},(x_{0},x_{1}+x_{2},x_{3}))$
	$\displaystyle\zeta_{2}(x_{0},x_{1},x_{2},x_{3})$	$\displaystyle=(x_{3}+x_{4},(x_{0},x_{1},x_{2}+x_{3})).$

We saw above that $[0,1]\times\Delta_{1}$ is the union of the triangles $\zeta_{0}(\Delta_{2})$ and $\zeta_{1}(\Delta_{2})$ , which fit together nicely along one edge. In the same way, it can be shown that $[0,1]\times\Delta_{2}$ is the union of the images of the maps $\zeta_{i}\colon\Delta_{3}\to[0,1]\times\Delta_{2}$ , and the intersection of any two of these images is another simplex of lower dimension. However, we will not need this so we omit the proof.

Interactive demo

When $k=3$ we have

	$\displaystyle\zeta_{0}(x_{0},x_{1},x_{2},x_{3},x_{4})$	$\displaystyle=(x_{1}+x_{2}+x_{3}+x_{4},(x_{0}+x_{1},x_{2},x_{3},x_{4}))$
	$\displaystyle\zeta_{1}(x_{0},x_{1},x_{2},x_{3},x_{4})$	$\displaystyle=(x_{2}+x_{3}+x_{4},(x_{0},x_{1}+x_{2},x_{3},x_{4}))$
	$\displaystyle\zeta_{2}(x_{0},x_{1},x_{2},x_{3},x_{4})$	$\displaystyle=(x_{3}+x_{4},(x_{0},x_{1},x_{2}+x_{3},x_{4}))$
	$\displaystyle\zeta_{3}(x_{0},x_{1},x_{2},x_{3},x_{4})$	$\displaystyle=(x_{4},(x_{0},x_{1},x_{2},x_{3}+x_{4})).$

Definition 14.11.

Given $h\colon[0,1]\times X\to Y$ and $u\colon\Delta_{k}\to X$ as before, we put

\sigma_{k}(u)=\sum_{i=0}^{k}(-1)^{i}(h\circ(\operatorname{id}\times u)\circ% \zeta_{i})\in C_{k+1}(Y).

We extend this linearly to define $\sigma_{k}\colon C_{k}(X)\to C_{k+1}(Y)$ .

Lemma 14.12.

(a)

Suppose that $0\leq i\leq k$ and $0\leq j\leq k+1$ , so we can form the composite

$\Delta_{k}\xrightarrow{\delta_{j}}\Delta_{k+1}\xrightarrow{\zeta_{i}}[0,1]% \times\Delta_{k}.$

If $j<i$ then $\zeta_{i}\delta_{j}$ is the same as the composite

$\Delta_{k}\xrightarrow{\zeta_{i-1}}[0,1]\times\Delta_{k-1}\xrightarrow{% \operatorname{id}\times\delta_{j}}[0,1]\times\Delta_{k},$

or in other words $\zeta_{i}\delta_{j}=(\operatorname{id}\times\delta_{j})\zeta_{i-1}$ .
(b)

On the other hand, if $j\geq i+2$ then $\zeta_{i}\delta_{j}=(\operatorname{id}\times\delta_{j-1})\zeta_{i}$ .
(c)

For $1\leq i\leq k$ we also have $\zeta_{i}\delta_{i}=\zeta_{i-1}\delta_{i}$ .
(d)

Finally, we have $\zeta_{0}\delta_{0}(x)=(1,x)$ and $\zeta_{k}\delta_{k+1}(x)=(0,x)$ .

Proof.

All the maps under discussion are affine maps from $\Delta_{k}$ to $[0,1]\times\Delta_{k}$ . To check that two such maps agree, it suffices to check that they have the same effect on the vertices of $\Delta_{k}$ . The map $\delta_{j}$ sends $e_{p}$ to $e_{p}$ (if $p<j$ ) or $e_{p+1}$ (if $p\geq j$ ). The map $\zeta_{i}$ sends $e_{p}$ to $(0,e_{p})$ (if $p\leq i$ ) or $(1,e_{p-1})$ (if $p>i$ ).

(a)

When $j<i$ we find that both $\zeta_{i}\delta_{j}$ and $(\operatorname{id}\times\delta_{j})\zeta_{i-1}$ have the following effect:

$e_{p}\mapsto\begin{cases}(0,e_{p})&\text{ if }p<j\\ (0,e_{p+1})&\text{ if }j\leq p<i\\ (1,e_{p})&\text{ if }i\leq p.\end{cases}$
(b)

When $j\geq i+2$ we find that both $\zeta_{i}\delta_{j}$ and $(\operatorname{id}\times\delta_{j-1})\zeta_{i}$ have the following effect:

$e_{p}\mapsto\begin{cases}(0,e_{p})&\text{ if }p\leq i\\ (1,e_{p-1})&\text{ if }i<p<j\\ (1,e_{p})&\text{ if }j\leq p.\end{cases}$
(c)

When $1\geq i\leq k$ we find that both $\zeta_{i}\delta_{i}$ and $\zeta_{i-1}\delta_{i}$ have the following effect:

$e_{p}\mapsto\begin{cases}(0,e_{p})&\text{ if }p<i\\ (1,e_{p})&\text{ if }i\leq p.\end{cases}$
(d)

We also have $\zeta_{0}\delta_{0}(e_{p})=\zeta_{0}(e_{p+1})=(1,e_{p})$ and $\zeta_{k}\delta_{k+1}(e_{p})=\zeta_{k}(e_{p})=(0,e_{p})$

∎

Video (Proposition 14.13 to Proposition 14.17)

Proposition 14.13.

For all $u\in C_{k}(X)$ we have

\partial(\sigma_{k}(u))+\sigma_{k-1}(\partial(u))=g_{\#}(u)-f_{\#}(u).

Proof.

It will be enough to prove this when $u\in S_{k}(X)\subset C_{k}(X)$ , so $u\colon\Delta_{k}\to X$ . We can then define $m=h\circ(\operatorname{id}\times u)\colon[0,1]\times\Delta_{k}\to X$ , so $\sigma_{k}(u)=\sum_{i=0}^{k}(-1)^{i}m\zeta_{i}$ . This gives

\partial(\sigma_{k}(u))=\sum_{i=0}^{k}\sum_{j=0}^{k+1}(-1)^{i+j}m\zeta_{i}% \delta_{j}.

We can divide this sum into four parts:

•

$A$ is the sum of the terms where $j<i$
•

$B$ is the sum of the terms where $i+2\leq j$
•

$C$ is the sum of the terms where $j=i$ with $1\leq i\leq k$
•

$D$ is the sum of the terms where $j=i+1$ with $0\leq i<k$
•

$E$ consists of the terms with $(i,j)=(0,0)$ or $(i,j)=(k,k+1)$ .

We thus have $\partial(\sigma_{k}(u))=A+B+C+D+E$ .

Now note that $\partial(u)=\sum_{q=0}^{k}(-1)^{q}u\delta_{q}$ , and

\sigma_{k-1}(u\delta_{q})=\sum_{p=0}^{k}(-1)^{p}(h\circ(\operatorname{id}% \times u\delta_{q})\circ\zeta_{p})=\sum_{p=0}^{k}(-1)^{p}m(\operatorname{id}% \times\delta_{q})\zeta_{p}.

This gives

\sigma_{k-1}(\partial(u))=\sum_{p=0}^{k}\sum_{q=0}^{k}(-1)^{p+q}m(% \operatorname{id}\times\delta_{q})\zeta_{p}.

We let $A^{\prime}$ be the sum of the terms where $q\leq p$ , and we let $B^{\prime}$ be the sum of the terms where $p<q$ .

Each term $(-1)^{i+j}m\zeta_{i}\delta_{j}$ in $A$ can be rewritten (using part (a) of the lemma) as $(-1)^{i+j}m(\operatorname{id}\times\delta_{j})\zeta_{i-1}$ . This can then be written as $-(-1)^{p+q}m(\operatorname{id}\times\delta_{q})\delta_{p}$ , where $p=i-1$ and $q=j$ . Because $j<i$ for terms in $A$ , we see that $q\leq p$ , so the rewritten term is the negative of a term in $A^{\prime}$ . Similarly, each term $(-1)^{i+j}m\zeta_{i}\delta_{j}$ in $B$ can be rewritten (using part (b) of the lemma) as $(-1)^{i+j}m(\operatorname{id}\times\delta_{j-1})\zeta_{i}$ . This can then be written as $-(-1)^{p+q}m(\operatorname{id}\times\delta_{q})\delta_{p}$ , where $p=i$ and $q=j-1$ . Because $j\geq i+2$ for terms in $B$ , we see that $q>p$ , so the rewritten term is the negative of a term in $B^{\prime}$ . Using this we see that $A^{\prime}=-A$ and $B^{\prime}=-B$ . A similar argument with part (c) of the lemma shows that $D=-C$ . We now have

	$\displaystyle\partial(\sigma_{k}(u))+\sigma_{k-1}(\partial(u))$	$\displaystyle=(A+B+C+D+E)+(A^{\prime}+B^{\prime})$
		$\displaystyle=(A+B+C-C+E)+(-A-B)=E.$

Also, part (d) of the lemma gives

	$\displaystyle m\zeta_{0}\delta_{0}(x)$	$\displaystyle=m(1,x)=h(1,u(x))=g(u(x))$
	$\displaystyle m\zeta_{k}\delta_{k+1}(x)$	$\displaystyle=m(0,x)=h(0,u(x))=f(u(x)).$

The first of these terms has sign $(-1)^{0+0}=+1$ , and the second has sign $(-1)^{k+(k+1)}=-1$ . We therefore have $E=g_{\#}(u)-f_{\#}(u)$ as required. ∎

We can gain some insight into the above proof by considering a simple special case. Suppose that $X={\mathbb{R}}^{N}$ and $Y={\mathbb{R}}^{M}$ . Suppose that $f\colon X\to Y$ is affine, i.e. $f(x)=Ax+b$ for some matrix $A$ and vector $b$ . Suppose that $g\colon X\to Y$ is also affine, and that $h$ is just the linear homotopy $h(t,x)=(1-t)f(x)+tg(x)$ . Consider a linear $3$ -simplex $u=\langle a_{0},a_{1},a_{2},a_{3}\rangle\in C_{3}(X)$ . As all the maps involved are affine, we see that

	$\displaystyle f_{*}(u)$	$\displaystyle=\langle f(a_{0}),f(a_{1}),f(a_{2}),f(a_{3})\rangle$
	$\displaystyle g_{*}(u)$	$\displaystyle=\langle g(a_{0}),g(a_{1}),g(a_{2}),g(a_{3})\rangle.$

As in Example 10.17, we will use abbreviated notation, writing $i$ for $a_{i}$ or $f(a_{i})$ , and $\overline{i}$ for $g(a_{i})$ , so the above equations become $f_{*}(0123)=0123$ and $g_{*}(0123)=\overline{0123}$ . It is then not hard to check that

h\circ(\operatorname{id}\times u)\circ\zeta_{2}=\langle f(a_{0}),f(a_{1}),f(a_% {2}),g(a_{2}),g(a_{3})\rangle=012\overline{23},

and similarly for the other terms in $\sigma(u)$ . The terms in $\partial(u)$ , $\sigma(u)$ , $\partial\sigma(u)$ and $\sigma\partial(u)$ can now be laid out as follows:

Most terms cancel in the indicated groups, which correspond to the expressions $A,\dotsc,E$ in the proof of Proposition 14.13, leaving $\partial\sigma(u)+\sigma\partial(u)=\overline{0123}-0123=g_{*}(u)-f_{*}(u)$ as expected. Here we have just displayed the case $k=3$ , but the pattern generalises in an obvious way to other values of $k$ . This presentation is only directly relevant for linear simplices and affine maps. However, in the general case, most of the work involves linear simplices in the space $[0,1]\times\Delta_{k}\subseteq{\mathbb{R}}^{k+2}$ , and then we finish up by applying the map $h\circ(\operatorname{id}\times u)\colon[0,1]\times\Delta_{k}\to Y$ . Because of this, it is possible to deduce the general case from the linear case, although we will not spell out the details here.

Corollary 14.14.

If $f\colon X\to Y$ is a homotopy equivalence, then $f_{*}\colon H_{*}(X)\to H_{*}(Y)$ is an isomorphism.

Proof.

Choose a map $g\colon Y\to X$ which is homotopy inverse to $f$ , so $g\circ f$ is homotopic to $\operatorname{id}_{X}$ and $f\circ g$ is homotopic to $\operatorname{id}_{Y}$ . As homology is a functor, the composite $g_{*}\circ f_{*}\colon H_{*}(X)\to H_{*}(X)$ is the same as $(g\circ f)_{*}$ . As $g\circ f$ is homotopic to $\operatorname{id}_{X}$ , Proposition 14.8 tells us that $(g\circ f)_{*}=(\operatorname{id}_{X})_{*}$ . Using functoriality again, we have $(\operatorname{id}_{X})_{*}=\operatorname{id}_{H_{*}(X)}$ . Putting this together, we see that $g_{*}\circ f_{*}=\operatorname{id}\colon H_{*}(X)\to H_{*}(X)$ , and essentially the same argument shows that $f_{*}\circ g_{*}=\operatorname{id}\colon H_{*}(Y)\to H_{*}(Y)$ . Thus, $f_{*}$ and $g_{*}$ are mutually inverse isomorphisms. ∎

Remark 14.15.

Another way to organise the above argument is as follows. Propositions 14.7 and 14.8 tell us that $H_{n}$ can be regarded as a functor $\operatorname{hTop}\to\operatorname{Ab}$ . Any homotopy equivalence $f\colon X\to Y$ becomes an isomorphism in $\operatorname{hTop}$ , and Corollary 6.18 tells us that functors send isomorphisms to isomorphisms, so $H_{n}(f)$ must be an isomorphism.

Proposition 14.16.

If $f\colon X\to Y$ is homotopic to a constant map, then the map $f_{*}\colon H_{n}(X)\to H_{n}(Y)$ is zero for all $n>0$ .

Proof.

Let $g\colon X\to Y$ be a constant map that is homotopic to $f$ . Then $f_{*}=g_{*}$ by Proposition 14.8, but $g_{*}=0$ by Remark 13.14. ∎

Proposition 14.17.

Suppose that $X$ is contractible. Then $H_{0}(X)={\mathbb{Z}}$ but $H_{n}(X)=0$ for all $n\neq 0$ . In particular, this applies if $X$ is a convex subset of ${\mathbb{R}}^{N}$ for some $n$ .

Proof.

Proposition 9.17 tells us that $X$ is homotopy equivalent to a point, so it has the same homology as a point, which is given by Proposition 10.23. ∎