Given a topological space $X$, we previously defined a system of groups ${C}_{k}(X)=\mathbb{Z}\{{S}_{k}(X)\}$ and homomorphisms $\partial :{C}_{k}(X)\to {C}_{k-1}(X)$. 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 ${H}_{k}(X)$.
A chain complex is a sequence of abelian groups and homomorphisms like
$$\mathrm{\cdots}\stackrel{}{\leftarrow}{A}_{-2}\stackrel{{d}_{-1}}{\leftarrow}{A}_{-1}\stackrel{{d}_{0}}{\leftarrow}{A}_{0}\stackrel{{d}_{1}}{\leftarrow}{A}_{1}\stackrel{{d}_{2}}{\leftarrow}{A}_{2}\stackrel{{d}_{3}}{\leftarrow}\mathrm{\cdots},$$ |
such that ${d}_{i}\circ {d}_{i+1}=0:{A}_{i+1}\to {A}_{i-1}$ 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
${Z}_{i}(A)$ | $=\mathrm{ker}({d}_{i}:{A}_{i}\to {A}_{i-1})\le {A}_{i}$ | ||
${B}_{i}(A)$ | $=\mathrm{img}({d}_{i+1}:{A}_{i+1}\to {A}_{i})\le {A}_{i}.$ |
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 ${B}_{i}(A)\le {Z}_{i}(A)$, 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.
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 ${H}_{*}(A)$.
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 ${C}_{*}(X)$.
Note that the quotient ${H}_{i}(A)={Z}_{i}(A)/{B}_{i}(A)$ is zero iff ${Z}_{i}(A)={B}_{i}(A)$ iff $\mathrm{ker}({d}_{i}:{A}_{i}\to {A}_{i-1})=\mathrm{img}({d}_{i+1}:{A}_{i+1}\to {A}_{i})$ iff the sequence ${A}_{i+1}\stackrel{{d}_{i+1}}{\to}{A}_{i}\stackrel{{d}_{i}}{\to}{A}_{i-1}$ 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 ${A}_{i}$.
Consider a chain complex in which ${A}_{i}=0$ for $i>0$, so the complex just has the form
$$\stackrel{}{\leftarrow}0\stackrel{{d}_{0}}{\leftarrow}{A}_{0}\stackrel{{d}_{1}}{\leftarrow}0\stackrel{{d}_{2}}{\leftarrow}0\stackrel{{d}_{3}}{\leftarrow}\mathrm{\cdots}$$ |
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 $i\ne 0$ we just have ${Z}_{i}(A)=0$ and ${B}_{i}(A)=0$ and ${H}_{i}(A)=0$.
Now consider a chain complex in which ${A}_{i}=0$ for $i>1$, so the complex just has the form
$$\stackrel{}{\leftarrow}0\stackrel{{d}_{0}}{\leftarrow}{A}_{0}\stackrel{{d}_{1}}{\leftarrow}{A}_{1}\stackrel{{d}_{2}}{\leftarrow}0\stackrel{{d}_{3}}{\leftarrow}\mathrm{\cdots}$$ |
It is clear that ${d}_{i}=0$ for all $i\ne 1$, but ${d}_{1}$ may be nonzero. It follows that
${Z}_{0}(A)$ | $={A}_{0}$ | ${B}_{0}(A)$ | $={d}_{1}({A}_{1})$ | ${H}_{0}(A)$ | $={A}_{0}/{d}_{1}({A}_{1})$ | ||
${Z}_{1}(A)$ | $=\mathrm{ker}({d}_{1})$ | ${B}_{1}(A)$ | $=0$ | ${H}_{1}(A)$ | $=\mathrm{ker}({d}_{1}).$ |
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:
${A}_{0}$ | $=\mathbb{Z}\{{v}_{0},\mathrm{\dots},{v}_{n-1}\}$ | ||
${A}_{1}$ | $=\mathbb{Z}\{{e}_{0},\mathrm{\dots},{e}_{n-1}\}.$ |
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 ${d}_{1}:{A}_{1}\to {A}_{0}$ 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
${e}_{i}^{\prime}$ | $$ | ||
${v}_{0}^{\prime}$ | $={v}_{0}$ | ||
${v}_{i}^{\prime}$ | $$ |
For example, when $n=4$ we have
${e}_{0}^{\prime}$ | $={e}_{0}$ | ${e}_{0}$ | $={e}_{0}^{\prime}$ | ${v}_{0}^{\prime}$ | $={v}_{0}$ | ${v}_{0}$ | $={v}_{0}^{\prime}$ | ||
${e}_{1}^{\prime}$ | $={e}_{0}+{e}_{1}$ | ${e}_{1}$ | $={e}_{1}^{\prime}-{e}_{0}^{\prime}$ | ${v}_{1}^{\prime}$ | $={v}_{1}-{v}_{0}$ | ${v}_{1}$ | $={v}_{1}^{\prime}+{v}_{0}^{\prime}$ | ||
${e}_{2}^{\prime}$ | $={e}_{0}+{e}_{1}+{e}_{2}$ | ${e}_{2}$ | $={e}_{2}^{\prime}-{e}_{1}^{\prime}$ | ${v}_{2}^{\prime}$ | $={v}_{2}-{v}_{0}$ | ${v}_{2}$ | $={v}_{2}^{\prime}+{v}_{0}^{\prime}$ | ||
${e}_{3}^{\prime}$ | $={e}_{0}+{e}_{1}+{e}_{2}+{e}_{3}$ | ${e}_{3}$ | $={e}_{3}^{\prime}-{e}_{2}^{\prime}$ | ${v}_{3}^{\prime}$ | $={v}_{3}-{v}_{0}$ | ${v}_{3}$ | $={v}_{3}^{\prime}+{v}_{0}^{\prime}$ |
It is easy to see that the list $({e}_{0}^{\prime},\mathrm{\dots},{e}_{n-1}^{\prime})$ is a basis for ${A}_{1}$ over $\mathbb{Z}$, and $({v}_{0}^{\prime},\mathrm{\dots},{v}_{n-1}^{\prime})$ is a basis for ${A}_{0}$. We also have
$${d}_{1}({e}_{i}^{\prime})=({v}_{1}-{v}_{0})+({v}_{2}-{v}_{1})+\mathrm{\cdots}+({v}_{i+1}-{v}_{i}).$$ |
Most of the terms cancel, giving ${d}_{1}({e}_{i}^{\prime})={v}_{i+1}-{v}_{0}$. For $$ we can write this as ${d}_{1}({e}_{i}^{\prime})={v}_{i+1}^{\prime}$. However, ${v}_{n}$ is the same as ${v}_{0}$, so ${d}_{1}^{\prime}({e}_{n-1}^{\prime})=0$. In summary, we have
$${d}_{1}({e}_{0}^{\prime})={v}_{1}^{\prime}\mathit{\hspace{1em}\hspace{1em}}{d}_{1}({e}_{1}^{\prime})={v}_{2}^{\prime}\mathit{\hspace{1em}\hspace{1em}}{d}_{1}({e}_{2}^{\prime})={v}_{3}^{\prime}\mathit{\hspace{1em}\hspace{1em}}\mathrm{\cdots}\mathit{\hspace{1em}\hspace{1em}}{d}_{1}({e}_{n-2}^{\prime})={v}_{n-1}^{\prime}\mathit{\hspace{1em}\hspace{1em}}{d}_{1}({e}_{n-1}^{\prime})=0.$$ |
From this it is clear that $\mathrm{ker}({d}_{1})=\mathbb{Z}.{e}_{n-1}^{\prime}$ and $\mathrm{img}({d}_{1})=\mathbb{Z}\{{v}_{1}^{\prime},\mathrm{\dots},{v}_{n-1}^{\prime}\}$ so ${A}_{0}/\mathrm{img}({d}_{1})=\mathbb{Z}.{v}_{0}^{\prime}$. In other words, we have ${H}_{1}(A)=\mathbb{Z}.{e}_{n-1}^{\prime}$ and ${H}_{0}(A)=\mathbb{Z}.{v}_{0}^{\prime}$.
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.
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.
Let $A$ and ${A}^{\prime}$ be chain complexes. A chain map from $A$ to ${A}^{\prime}$ is a sequence of group homomorphisms ${f}_{n}:{A}_{n}\to {A}_{n}^{\prime}$ such that for all $n\ge 0$ we have ${f}_{n}\circ {d}_{n+1}^{A}={d}_{n+1}^{{A}^{\prime}}\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}_{n+1}^{A}={d}_{n+1}^{{A}^{\prime}}\circ {f}_{n+1}$.
We write ${\mathrm{id}}_{{A}_{*}}$ for the sequence of identity maps ${\mathrm{id}}_{{A}_{n}}:{A}_{n}\to {A}_{n}$. This is clearly a chain map from $A$ to itself. Now suppose we have chain maps $A\stackrel{\mathit{f}}{\to}{A}^{\prime}\stackrel{{f}^{\prime}}{\to}{A}^{\prime \prime}$. We define a chain map ${f}^{\prime}\circ f:A\to {A}^{\prime \prime}$ by the obvious rule ${({f}^{\prime}\circ f)}_{n}={f}_{n}^{\prime}\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$$ |
as required. It is easy to see that this kind of composition is associative and that $\mathrm{id}\circ f=f=f\circ \mathrm{id}$, so we have a category $\mathrm{Chain}$ of chain complexes and chain maps.
Let $f\mathrm{:}A\mathrm{\to}{A}^{\mathrm{\prime}}$ be a chain map. Then
For all $n$ we have ${f}_{n}({Z}_{n}(A))\le {Z}_{n}({A}^{\prime})$
For all $n$ we have ${f}_{n}({B}_{n}(A))\le {B}_{n}({A}^{\prime})$
There is a well-defined map ${f}_{*}:{H}_{n}(A)\to {H}_{n}({A}^{\prime})$ given by ${f}_{*}[z]=[f(z)]$.
Moreover, these constructions give functors ${Z}_{n}\mathrm{,}{B}_{n}\mathrm{,}{H}_{n}\mathrm{:}\mathrm{Chain}\mathrm{\to}\mathrm{Ab}$.
Suppose that $z\in {Z}_{n}(A)$, so $d(z)=0$. We then have $f(z)\in {A}_{n}^{\prime}$ with $d(f(z))=f(d(z))=f(0)=0$, so $f(z)\in {Z}_{n}({A}^{\prime})$ as required. (As mentioned in Remark 10.22, the notation $[{z}^{\prime}]$ is only meaningful if $d({z}^{\prime})=0$. Now we have checked that $d(f(z))=0$, we see that the notation $[f(z)]$ used in (c) is valid.)
Suppose that $b\in {B}_{n}(A)$, so $b=d(x)$ for some $x\in {A}_{n+1}$. We then have $f(b)=fd(x)=df(x)\in d({A}_{n+1}^{\prime})={B}_{n}({A}^{\prime})$ as required.
From (a) and (b) we see that if $z+{B}_{n}(A)={z}^{\prime}+{B}_{n}(A)$ then $z-{z}^{\prime}\in {B}_{n}(A)$ so $f(z)-f({z}^{\prime})=f(z-{z}^{\prime})\in {B}_{n}({A}^{\prime})$ so $f(z)+{B}_{n}({A}^{\prime})=f({z}^{\prime})+{B}_{n}({A}^{\prime})$. It follows that there is an induced homomorphism ${f}_{*}:{H}_{n}(A)\to {H}_{n}({A}^{\prime})$ given by
$${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)}:{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\stackrel{\mathit{f}}{\to}{A}^{\prime}\stackrel{{f}^{\prime}}{\to}{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 ${H}_{n}:\mathrm{Chain}\to \mathrm{Ab}$ is also a functor. ∎
We now want to make homology into a functor from topological spaces to abelian groups. Let $f:X\to Y$ be a continuous map. Let $u$ be an element of the set ${S}_{k}(X)$, or equivalently, a continuous map $u:{\mathrm{\Delta}}_{k}\to X$. We define ${f}_{\mathrm{\#}}(u)\in {S}_{k}(Y)$ to be the composite function $f\circ u:{\mathrm{\Delta}}_{k}\to Y$. Next, given an element $u={\sum}_{i=1}^{r}{n}_{i}{u}_{i}\in {C}_{k}(X)$ we define ${f}_{\mathrm{\#}}(u)={\sum}_{i=1}^{r}{n}_{i}{f}_{\mathrm{\#}}({u}_{i})\in {C}_{k}(Y)$. This extends the map ${f}_{\mathrm{\#}}:{S}_{k}(X)\to {S}_{k}(Y)$ linearly to give a homomorphism ${f}_{\mathrm{\#}}:{C}_{k}(X)\to {C}_{k}(Y)$. We claim that this is a chain map, or in other words that ${f}_{\mathrm{\#}}(\partial (u))=\partial ({f}_{\mathrm{\#}}(u))$ 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 $u:{\mathrm{\Delta}}_{k}\to X$. We then have
${f}_{\mathrm{\#}}(u)$ | $=f\circ u$ | ||
$\partial (u)$ | $={\displaystyle \sum _{i=0}^{k}}{(-1)}^{i}(u\circ {\delta}_{i})$ | ||
${f}_{\mathrm{\#}}(\partial (u))$ | $={\displaystyle \sum _{i=0}^{k}}{(-1)}^{i}(f\circ u\circ {\delta}_{i})=\partial ({f}_{\mathrm{\#}}(u)),$ |
as required. It is clear that ${\mathrm{id}}_{\mathrm{\#}}=\mathrm{id}$ and ${(g\circ f)}_{\mathrm{\#}}={g}_{\mathrm{\#}}\circ {f}_{\mathrm{\#}}$, so we have defined a functor ${C}_{*}:\mathrm{Top}\to \mathrm{Chain}$. We can compose this with the functor ${H}_{n}:\mathrm{Chain}\to \mathrm{Ab}$ to get a functor $\mathrm{Top}\to \mathrm{Ab}$ which is traditionally also denoted by ${H}_{n}$.
Suppose that $f:X\to Y$ is constant, say $f(x)=b$ for all $x\in X$. We then claim that ${f}_{*}=0:{H}_{n}(X)\to {H}_{n}(Y)$ for all $n>0$. Indeed, we can define $X\stackrel{\mathit{p}}{\to}\{0\}\stackrel{\mathit{q}}{\to}Y$ by $p(x)=0$ and $q(0)=b$, so $f=q\circ p$. Thus, the map ${f}_{*}:{H}_{n}(X)\to {H}_{n}(Y)$ is the composite of the maps ${p}_{*}:{H}_{n}(X)\to {H}_{n}(\{0\})$ and ${q}_{*}:{H}_{n}(\{0\})\to {H}_{n}(Y)$. However, Proposition 10.23 tells us that ${H}_{n}(\{0\})=0$, and the claim is clear from this.