MAS61015 Algebraic Topology 12 Abelian groups 14 Chain homotopy

13. Chain complexes and homology

Video (Definition 13.1 to Example 13.7)

Given a topological space $X$ , we previously defined a system of groups $C_{k}(X)={\mathbb{Z}}\{S_{k}(X)\}$ and homomorphisms $\partial\colon C_{k}(X)\to C_{k-1}(X)$ . In Proposition 10.16 we showed that $\partial^{2}=0$ , and this allowed us to define the homology groups $H_{k}(X)$ . In this section we will place all those constructions in a wider context, which will be useful for calculating $H_{k}(X)$ .

Definition 13.1.

A chain complex is a sequence of abelian groups and homomorphisms like

\dotsb\xleftarrow{}A_{-2}\xleftarrow{d_{-1}}A_{-1}\xleftarrow{d_{0}}A_{0}% \xleftarrow{d_{1}}A_{1}\xleftarrow{d_{2}}A_{2}\xleftarrow{d_{3}}\dotsb,

such that $d_{i}\circ d_{i+1}=0\colon A_{i+1}\to A_{i-1}$ for all $i$ . We will often suppress the indices and just write $d^{2}=0$ instead of $d_{i}\circ d_{i+1}=0$ . We now put

	$\displaystyle Z_{i}(A)$	$\displaystyle=\ker(d_{i}\colon A_{i}\to A_{i-1})\leq A_{i}$
	$\displaystyle B_{i}(A)$	$\displaystyle=\operatorname{img}(d_{i+1}\colon A_{i+1}\to A_{i})\leq A_{i}.$

The elements of $Z_{i}(A)$ are called cycles, and the elements of $B_{i}(A)$ are called boundaries. As in Remark 12.13, the condition $d_{i}\circ d_{i+1}=0$ means that $B_{i}(A)\leq Z_{i}(A)$ , so every boundary is a cycle. Because of this, it is legitimate to define

H_{i}(A)=Z_{i}(A)/B_{i}(A).

This is called the $i$ ’th homology group of the complex. Given a cycle $z\in Z_{i}(A)$ , we write $[z]$ for the coset $z+B_{i}(A)\in H_{i}(A)$ , and call this the homology class of $z$ . Thus $[z]$ is defined iff $dz=0$ , and $[z]=[z^{\prime}]$ iff $z^{\prime}=z+dy$ for some $y$ . The homomorphisms $d_{i}$ are called differentials.

Remark 13.2.

We use the notation $A_{*}$ to refer to the whole chain complex as a single object. We also write $Z_{*}(A)$ for the whole system of groups $Z_{i}(A)$ , and similarly for $B_{*}(A)$ and $H_{*}(A)$ .

Remark 13.3.

The content of Section 10 can now be expressed as follows: the groups $C_{*}(X)$ form a chain complex (with differential $\partial$ ), and the homology groups of the space $X$ are defined to be the homology groups of the chain complex $C_{*}(X)$ .

Remark 13.4.

Note that the quotient $H_{i}(A)=Z_{i}(A)/B_{i}(A)$ is zero iff $Z_{i}(A)=B_{i}(A)$ iff $\ker(d_{i}\colon A_{i}\to A_{i-1})=\operatorname{img}(d_{i+1}\colon A_{i+1}\to A% _{i})$ iff the sequence $A_{i+1}\xrightarrow{d_{i+1}}A_{i}\xrightarrow{d_{i}}A_{i-1}$ is exact. Thus, the size of the group $H_{i}(A)$ can be regarded as a measure of how badly the chain complex fails to be exact at $A_{i}$ .

Example 13.5.

Consider a chain complex in which $A_{i}=0$ for $i>0$ , so the complex just has the form

\xleftarrow{}0\xleftarrow{d_{0}}A_{0}\xleftarrow{d_{1}}0\xleftarrow{d_{2}}0% \xleftarrow{d_{3}}\dotsb

Every differential either starts or ends at the zero group, so all differentials are zero. We find that $Z_{0}(A)=A_{0}$ and $B_{0}(A)=0$ so $H_{0}(A)=A_{0}/0=A_{0}$ . For $i\neq 0$ we just have $Z_{i}(A)=0$ and $B_{i}(A)=0$ and $H_{i}(A)=0$ .

Example 13.6.

Now consider a chain complex in which $A_{i}=0$ for $i>1$ , so the complex just has the form

\xleftarrow{}0\xleftarrow{d_{0}}A_{0}\xleftarrow{d_{1}}A_{1}\xleftarrow{d_{2}}% 0\xleftarrow{d_{3}}\dotsb

It is clear that $d_{i}=0$ for all $i\neq 1$ , but $d_{1}$ may be nonzero. It follows that

	$\displaystyle Z_{0}(A)$	$\displaystyle=A_{0}$	$\displaystyle B_{0}(A)$	$\displaystyle=d_{1}(A_{1})$	$\displaystyle H_{0}(A)$	$\displaystyle=A_{0}/d_{1}(A_{1})$
	$\displaystyle Z_{1}(A)$	$\displaystyle=\ker(d_{1})$	$\displaystyle B_{1}(A)$	$\displaystyle=0$	$\displaystyle H_{1}(A)$	$\displaystyle=\ker(d_{1}).$

Example 13.7.

We now consider a specific chain complex $A$ of the type discussed in Example 13.6, so $A_{k}=0$ for $k>1$ . We fix $n>0$ , and take $A_{0}$ and $A_{1}$ to be free abelian groups as follows:

	$\displaystyle A_{0}$	$\displaystyle={\mathbb{Z}}\{v_{0},\dotsc,v_{n-1}\}$
	$\displaystyle A_{1}$	$\displaystyle={\mathbb{Z}}\{e_{0},\dotsc,e_{n-1}\}.$

These indices are supposed to be read modulo $n$ , so $v_{n}=v_{0}$ and $v_{-1}=v_{n-1}$ and so on. We define $d_{1}\colon A_{1}\to A_{0}$ by $d_{1}(e_{i})=v_{i+1}-v_{i}$ (so in particular $d_{1}(e_{n-1})=v_{0}-v_{n-1}$ ). We claim that the groups $H_{0}(A)$ and $H_{1}(A)$ are both isomorphic to ${\mathbb{Z}}$ .

The easiest way to see this is to introduce some alternative bases for these groups. We put

	$\displaystyle e^{\prime}_{i}$	$\displaystyle=e_{0}+e_{1}+\dotsb+e_{i}=\sum_{j=0}^{i}e_{j}\qquad(\text{ for }0% \leq i<n)$
	$\displaystyle v^{\prime}_{0}$	$\displaystyle=v_{0}$
	$\displaystyle v^{\prime}_{i}$	$\displaystyle=v_{i}-v_{0}\qquad(\text{ for }0<i<n).$

For example, when $n=4$ we have

$\displaystyle e^{\prime}_{0}$	$\displaystyle=e_{0}$	$\displaystyle e_{0}$	$\displaystyle=e^{\prime}_{0}$	$\displaystyle v^{\prime}_{0}$	$\displaystyle=v_{0}$	$\displaystyle v_{0}$	$\displaystyle=v^{\prime}_{0}$
$\displaystyle e^{\prime}_{1}$	$\displaystyle=e_{0}+e_{1}$	$\displaystyle e_{1}$	$\displaystyle=e^{\prime}_{1}-e^{\prime}_{0}$	$\displaystyle v^{\prime}_{1}$	$\displaystyle=v_{1}-v_{0}$	$\displaystyle v_{1}$	$\displaystyle=v^{\prime}_{1}+v^{\prime}_{0}$
$\displaystyle e^{\prime}_{2}$	$\displaystyle=e_{0}+e_{1}+e_{2}$	$\displaystyle e_{2}$	$\displaystyle=e^{\prime}_{2}-e^{\prime}_{1}$	$\displaystyle v^{\prime}_{2}$	$\displaystyle=v_{2}-v_{0}$	$\displaystyle v_{2}$	$\displaystyle=v^{\prime}_{2}+v^{\prime}_{0}$
$\displaystyle e^{\prime}_{3}$	$\displaystyle=e_{0}+e_{1}+e_{2}+e_{3}$	$\displaystyle e_{3}$	$\displaystyle=e^{\prime}_{3}-e^{\prime}_{2}$	$\displaystyle v^{\prime}_{3}$	$\displaystyle=v_{3}-v_{0}$	$\displaystyle v_{3}$	$\displaystyle=v^{\prime}_{3}+v^{\prime}_{0}$

It is easy to see that the list $(e^{\prime}_{0},\dotsc,e^{\prime}_{n-1})$ is a basis for $A_{1}$ over ${\mathbb{Z}}$ , and $(v^{\prime}_{0},\dotsc,v^{\prime}_{n-1})$ is a basis for $A_{0}$ . We also have

d_{1}(e^{\prime}_{i})=(v_{1}-v_{0})+(v_{2}-v_{1})+\dotsb+(v_{i+1}-v_{i}).

Most of the terms cancel, giving $d_{1}(e^{\prime}_{i})=v_{i+1}-v_{0}$ . For $0\leq i<n-1$ we can write this as $d_{1}(e^{\prime}_{i})=v^{\prime}_{i+1}$ . However, $v_{n}$ is the same as $v_{0}$ , so $d^{\prime}_{1}(e^{\prime}_{n-1})=0$ . In summary, we have

d_{1}(e^{\prime}_{0})=v^{\prime}_{1}\qquad d_{1}(e^{\prime}_{1})=v^{\prime}_{2% }\qquad d_{1}(e^{\prime}_{2})=v^{\prime}_{3}\qquad\dotsb\qquad d_{1}(e^{\prime% }_{n-2})=v^{\prime}_{n-1}\qquad d_{1}(e^{\prime}_{n-1})=0.

From this it is clear that $\ker(d_{1})={\mathbb{Z}}.e^{\prime}_{n-1}$ and $\operatorname{img}(d_{1})={\mathbb{Z}}\{v^{\prime}_{1},\dotsc,v^{\prime}_{n-1}\}$ so $A_{0}/\operatorname{img}(d_{1})={\mathbb{Z}}.v^{\prime}_{0}$ . In other words, we have $H_{1}(A)={\mathbb{Z}}.e^{\prime}_{n-1}$ and $H_{0}(A)={\mathbb{Z}}.v^{\prime}_{0}$ .

We will typically express this answer by writing $H_{*}(A)=({\mathbb{Z}},{\mathbb{Z}})$ . The first group listed is $H_{0}$ , the second one is $H_{1}$ , and it is implicit that all subsequent groups are zero.

Remark 13.8.

The method used above is fairly typical for calculation of the homology of small chain complexes. To calculate $H_{*}(A)$ , we try to find bases for all the groups $A_{i}$ such that the differential sends every basis element to another basis element or to zero. We can then take the basis for $A$ and discard all pairs of basis elements $(a,a^{\prime})$ with $da=a^{\prime}$ . The remaining elements will then give a basis for $H_{*}(A)$ over ${\mathbb{Z}}$ , so in particular all the groups $H_{i}(A)$ are free abelian groups.

There are some chain complexes $A$ for which the groups $H_{i}(A)$ are not free abelian groups, and in those cases we will not be able to find a basis with properties as above. There is a more complicated algorithm that will still work in those cases, but we will not explain it here.

Video (Definition 13.9 to Remark 13.13)

Definition 13.9.

Let $A$ and $A^{\prime}$ be chain complexes. A chain map from $A$ to $A^{\prime}$ is a sequence of group homomorphisms $f_{n}\colon A_{n}\to A^{\prime}_{n}$ such that for all $n\geq 0$ we have $f_{n}\circ d^{A}_{n+1}=d^{A^{\prime}}_{n+1}\circ f_{n+1}$ . In other words, the following diagram must commute:

We will typically suppress all subscripts and superscripts and just write $fd=df$ rather than $f_{n}\circ d^{A}_{n+1}=d^{A^{\prime}}_{n+1}\circ f_{n+1}$ .

Definition 13.10.

We write $\operatorname{id}_{A_{*}}$ for the sequence of identity maps $\operatorname{id}_{A_{n}}\colon A_{n}\to A_{n}$ . This is clearly a chain map from $A$ to itself. Now suppose we have chain maps $A\xrightarrow{f}A^{\prime}\xrightarrow{f^{\prime}}A^{\prime\prime}$ . We define a chain map $f^{\prime}\circ f\colon A\to A^{\prime\prime}$ by the obvious rule $(f^{\prime}\circ f)_{n}=f^{\prime}_{n}\circ f_{n}$ ; this satisfies

d\circ(f^{\prime}\circ f)=d\circ f^{\prime}\circ f=f^{\prime}\circ d\circ f=(f% ^{\prime}\circ f)\circ d

as required. It is easy to see that this kind of composition is associative and that $\operatorname{id}\circ f=f=f\circ\operatorname{id}$ , so we have a category $\operatorname{Chain}$ of chain complexes and chain maps.

Proposition 13.11.

Let $f\colon A\to A^{\prime}$ be a chain map. Then

(a)

For all $n$ we have $f_{n}(Z_{n}(A))\leq Z_{n}(A^{\prime})$
(b)

For all $n$ we have $f_{n}(B_{n}(A))\leq B_{n}(A^{\prime})$
(c)

There is a well-defined map $f_{*}\colon H_{n}(A)\to H_{n}(A^{\prime})$ given by $f_{*}[z]=[f(z)]$ .

Moreover, these constructions give functors $Z_{n},B_{n},H_{n}\colon\operatorname{Chain}\to\operatorname{Ab}$ .

Proof.

(a)

Suppose that $z\in Z_{n}(A)$ , so $d(z)=0$ . We then have $f(z)\in A^{\prime}_{n}$ with $d(f(z))=f(d(z))=f(0)=0$ , so $f(z)\in Z_{n}(A^{\prime})$ as required. (As mentioned in Remark 10.22, the notation $[z^{\prime}]$ is only meaningful if $d(z^{\prime})=0$ . Now we have checked that $d(f(z))=0$ , we see that the notation $[f(z)]$ used in (c) is valid.)
(b)

Suppose that $b\in B_{n}(A)$ , so $b=d(x)$ for some $x\in A_{n+1}$ . We then have $f(b)=fd(x)=df(x)\in d(A^{\prime}_{n+1})=B_{n}(A^{\prime})$ as required.
(c)

From (a) and (b) we see that if $z+B_{n}(A)=z^{\prime}+B_{n}(A)$ then $z-z^{\prime}\in B_{n}(A)$ so $f(z)-f(z^{\prime})=f(z-z^{\prime})\in B_{n}(A^{\prime})$ so $f(z)+B_{n}(A^{\prime})=f(z^{\prime})+B_{n}(A^{\prime})$ . It follows that there is an induced homomorphism $f_{*}\colon H_{n}(A)\to H_{n}(A^{\prime})$ given by

$f_{*}(z+B_{n}(A))=f(z)+B_{n}(A^{\prime}),$

or in other words $f_{*}[z]=[f(z)]$ .

We can make $Z_{n}$ into a functor by defining

Z_{n}f=f_{n}|_{Z_{n}(A)}\colon Z_{n}(A)\to Z_{n}(A^{\prime}).

As this is just a restriction of $f_{n}$ , it is clearly compatible with composition and identity morphisms, as required. We can make $B_{n}$ into a functor in the same way. Finally, suppose we have chain maps

A\xrightarrow{f}A^{\prime}\xrightarrow{f^{\prime}}A^{\prime\prime}.

For a homology class $[z]\in H_{n}(A)$ we have

(f^{\prime}\circ f)_{*}([z])=[(f^{\prime}\circ f)(z)]=[f^{\prime}(f(z))]=f^{% \prime}_{*}[f(z)]=f^{\prime}_{*}f_{*}[z].

From this it is clear that $H_{n}\colon\operatorname{Chain}\to\operatorname{Ab}$ is also a functor. ∎

Construction 13.12.

We now want to make homology into a functor from topological spaces to abelian groups. Let $f\colon X\to Y$ be a continuous map. Let $u$ be an element of the set $S_{k}(X)$ , or equivalently, a continuous map $u\colon\Delta_{k}\to X$ . We define $f_{\#}(u)\in S_{k}(Y)$ to be the composite function $f\circ u\colon\Delta_{k}\to Y$ . Next, given an element $u=\sum_{i=1}^{r}n_{i}u_{i}\in C_{k}(X)$ we define $f_{\#}(u)=\sum_{i=1}^{r}n_{i}f_{\#}(u_{i})\in C_{k}(Y)$ . This extends the map $f_{\#}\colon S_{k}(X)\to S_{k}(Y)$ linearly to give a homomorphism $f_{\#}\colon C_{k}(X)\to C_{k}(Y)$ . We claim that this is a chain map, or in other words that $f_{\#}(\partial(u))=\partial(f_{\#}(u))$ whenever $u\in C_{k}(X)$ . As everything is extended linearly, it will be enough to prove this when $u\in S_{k}(X)$ , or equivalently $u\colon\Delta_{k}\to X$ . We then have

	$\displaystyle f_{\#}(u)$	$\displaystyle=f\circ u$
	$\displaystyle\partial(u)$	$\displaystyle=\sum_{i=0}^{k}(-1)^{i}(u\circ\delta_{i})$
	$\displaystyle f_{\#}(\partial(u))$	$\displaystyle=\sum_{i=0}^{k}(-1)^{i}(f\circ u\circ\delta_{i})=\partial(f_{\#}(% u)),$

as required. It is clear that $\operatorname{id}_{\#}=\operatorname{id}$ and $(g\circ f)_{\#}=g_{\#}\circ f_{\#}$ , so we have defined a functor $C_{*}\colon\operatorname{Top}\to\operatorname{Chain}$ . We can compose this with the functor $H_{n}\colon\operatorname{Chain}\to\operatorname{Ab}$ to get a functor $\operatorname{Top}\to\operatorname{Ab}$ which is traditionally also denoted by $H_{n}$ .

Remark 13.13.

Now that we know that $H_{n}$ is a functor, we can use Propositions 6.17 and 6.25. These tell us that:

(a)

If $X$ is homeomorphic to $Y$ , then $H_{n}(X)$ is isomorphic to $H_{n}(Y)$ .
(b)

If $X$ is a retract of $Y$ , then $H_{n}(X)$ is a retract of $H_{n}(Y)$ .

Remark 13.14.

Suppose that $f\colon X\to Y$ is constant, say $f(x)=b$ for all $x\in X$ . We then claim that $f_{*}=0\colon H_{n}(X)\to H_{n}(Y)$ for all $n>0$ . Indeed, we can define $X\xrightarrow{p}\{0\}\xrightarrow{q}Y$ by $p(x)=0$ and $q(0)=b$ , so $f=q\circ p$ . Thus, the map $f_{*}\colon H_{n}(X)\to H_{n}(Y)$ is the composite of the maps $p_{*}\colon H_{n}(X)\to H_{n}(\{0\})$ and $q_{*}\colon H_{n}(\{0\})\to H_{n}(Y)$ . However, Proposition 10.23 tells us that $H_{n}(\{0\})=0$ , and the claim is clear from this.