Let $X$ be a space. A singular $k$-simplex in $X$ is a continuous map $u:{\mathrm{\Delta}}_{k}\to X$. We write ${S}_{k}(X)$ for the set of singular $k$-simplices in $X$.
Recall that ${\mathrm{\Delta}}_{0}$ is the set $\{{e}_{0}\}$ with just one point. To give a function $u:{\mathrm{\Delta}}_{0}\to X$ is the same as to give a point $u({e}_{0})\in X$; this, we can identify ${S}_{0}(X)$ with $X$.
As usual, we identify ${\mathrm{\Delta}}_{1}$ with $[0,1]$, with the point $(1-t,t)\in {\mathrm{\Delta}}_{1}$ corresponding to the point $t\in [0,1]$. Thus, a singular $1$-simplex in $X$ is the same as a continuous map $u:[0,1]\to X$, or in other words a path in $X$. This means that ${S}_{1}(X)$ is the set of all possible paths in $X$.
Suppose that ${a}_{0},\mathrm{\dots},{a}_{k}\in {\mathbb{R}}^{N}$. We can then define a map
$$\u27e8a\u27e9=\u27e8{a}_{0},\mathrm{\dots},{a}_{k}\u27e9:{\mathrm{\Delta}}_{k}\to {\mathbb{R}}^{N}$$ |
(or in other words an element $\u27e8a\u27e9\in {S}_{k}{\mathbb{R}}^{N}$) by
$$\u27e8a\u27e9({t}_{0},\mathrm{\dots},{t}_{n})={t}_{0}{a}_{0}+\mathrm{\cdots}+{t}_{k}{a}_{k}.$$ |
We call maps of this type linear simplices.
In the case $k=0$ we have ${S}_{0}(X)=X$ and the map $\u27e8{a}_{0}\u27e9$ just corresponds to the point ${a}_{0}$. In the case $k=1$, the map $\u27e8{a}_{0},{a}_{1}\u27e9$ corresponds to the straight line path from ${a}_{0}$ to ${a}_{1}$. In the case $k=2$, the image of the map $\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9:{\mathrm{\Delta}}_{2}\to {\mathbb{R}}^{N}$ is the triangle with vertices ${a}_{0}$, ${a}_{1}$ and ${a}_{2}$.
Now suppose that $X\subseteq {\mathbb{R}}^{N}$ and that ${a}_{0},\mathrm{\dots},{a}_{k}\in X$. It may or may not happen that the image of the map $\u27e8a\u27e9:{\mathrm{\Delta}}_{k}\to {\mathbb{R}}^{N}$ actually lies in $X$; this must be checked carefully in any context where we want to use this construction. If so, we can regard $\u27e8a\u27e9$ as an element of ${S}_{k}(X)$.
This picture shows a space $X\subset {\mathbb{R}}^{2}$, together with:
A linear $2$-simplex $\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9\in {S}_{2}{\mathbb{R}}^{2}$, which is not an element of ${S}_{2}(X)$.
Another linear $2$-simplex $\u27e8{b}_{0},{b}_{1},{b}_{2}\u27e9\in {S}_{2}(X)\subset {S}_{2}({\mathbb{R}}^{2})$.
A nonlinear $1$-simplex $u\in {S}_{1}(X)$.
Let $P$ be a set. We write $\mathbb{Z}\{P\}$ for the set of formal $\mathbb{Z}$-linear combinations of elements of $P$. Thus, if $p,q,r\in P$ then $5p-9q+7r\in \mathbb{Z}\{P\}$, for example. We call $\mathbb{Z}\{P\}$ the free abelian group generated by $P$. (It is clearly an abelian group under addition.)
Suppose that $P$ is finite, say $P=\{{p}_{1}\mathrm{\dots},{p}_{n}\}$. We then have an isomorphism $\varphi :{\mathbb{Z}}^{n}\to \mathbb{Z}\{P\}$ given by
$$\varphi ({a}_{1},\mathrm{\dots},{a}_{n})={a}_{1}{p}_{1}+\mathrm{\cdots}+{a}_{n}{p}_{n}.$$ |
However, we will most often be considering cases where $P$ is infinite.
A singular $k$-chain in $X$ is a formal $\mathbb{Z}$-linear combination of singular $k$-simplices, or in other words, an element of $\mathbb{Z}\{{S}_{k}(X)\}$. We write ${C}_{k}(X)=\mathbb{Z}\{{S}_{k}(X)\}$ for the group of singular $k$-chains. For convenience, we also define ${C}_{k}(X)=0$ for $$.
Consider again the picture in Example 10.5:
The expression $6{a}_{1}-4{b}_{2}+7{c}_{1}\in {C}_{0}(X)$ is a singular $0$-chain.
The expression $3\u27e8{a}_{0},{a}_{2}\u27e9-\u27e8{b}_{0},{b}_{1}\u27e9+u\in {C}_{1}(X)$ is a singular $1$-chain.
The expression $\u27e8{b}_{0},{b}_{1},{b}_{2}\u27e9\in {S}_{2}(X)\subset {C}_{2}(X)$ is a singular $2$-chain.
No expression involving $\u27e8{a}_{0},{a}_{1}\u27e9$ gives a singular chain in $X$, because the straight line from ${a}_{0}$ to ${a}_{1}$ is not contained in $X$.
Suppose we have paths $u:a\rightsquigarrow b$ and $v:b\rightsquigarrow c$ in $X$. We can reverse $u$ to get a path $\overline{u}:b\rightsquigarrow a$, or we can join $u$ and $v$ to get a path $u*v:a\rightsquigarrow c$.
We can regard $u*v$ and $u+v$ as elements of ${C}_{1}(X)$, but they are not the same. Similarly, we can regard $\overline{u}$ and $-u$ as elements of ${C}_{1}(X)$, but they are not the same. There is clearly an important relationship between $u*v$ and $u+v$, and between $\overline{u}$ and $-u$, but it will take a little work to formulate this mathematically.
We next need to define the algebraic boundary $\partial u\in {C}_{k-1}(X)$ for a $k$-chain $u\in {C}_{k}(X)$. We start by considering the cases $k=0$, $k=1$ and $k=2$.
For $u\in {C}_{0}(X)=\mathbb{Z}\{X\}$ we just define $\partial u=0$.
Now consider a singular $1$-simplex $u:{\mathrm{\Delta}}_{1}\to X$. This is a path with endpoints $u({e}_{0})$ and $u({e}_{1})$. These endpoints are elements of the set $X$, which we identify with ${S}_{0}(X)$, so the difference $u({e}_{1})-u({e}_{0})$ can be regarded as an element of ${C}_{0}(X)$. We define $\partial (u)=u({e}_{1})-u({e}_{0})$. More generally, suppose we have a $1$-chain $u={n}_{1}{u}_{1}+\mathrm{\cdots}+{n}_{r}{u}_{r}$, with ${u}_{i}:{\mathrm{\Delta}}_{1}\to X$ and ${n}_{i}\in \mathbb{Z}$. We then put
$$\partial (u)={n}_{1}\partial ({u}_{1})+\mathrm{\cdots}+{n}_{r}\partial ({u}_{r})=\sum _{i=1}^{r}{n}_{i}({u}_{i}({e}_{1})-{u}_{i}({e}_{0})).$$ |
This defines a homomorphism $\partial :{C}_{1}(X)\to {C}_{0}(X)$.
Note that for a linear $1$-simplex $\u27e8{a}_{0},{a}_{1}\u27e9$, we just have $\partial (\u27e8{a}_{0},{a}_{1}\u27e9)={a}_{1}-{a}_{0}$. Thus, in the picture below we have a $1$-chain
$$u=\u27e8{a}_{0},{a}_{1}\u27e9+\u27e8{a}_{1},{a}_{2}\u27e9+\u27e8{a}_{2},{a}_{3}\u27e9+\u27e8{a}_{3},{a}_{4}\u27e9+\u27e8{a}_{4},{a}_{5}\u27e9+\u27e8{a}_{5},{a}_{6}\u27e9$$ |
with
$$\partial (u)=({a}_{1}-{a}_{0})+({a}_{2}-{a}_{1})+({a}_{3}-{a}_{2})+({a}_{4}-{a}_{3})+({a}_{5}-{a}_{4})+({a}_{6}-{a}_{5})={a}_{6}-{a}_{0}.$$ |
We now consider $2$-chains. For the simplest case, suppose that $X\subseteq {\mathbb{R}}^{N}$ and $u=\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9$ is a linear $2$-simplex. In this case, we define
$$\partial (\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9)=\u27e8{a}_{1},{a}_{2}\u27e9-\u27e8{a}_{0},{a}_{2}\u27e9+\u27e8{a}_{0},{a}_{1}\u27e9.$$ |
The rule for nonlinear singular $2$-simplices is essentially a straightforward adaptation of the linear case, but it will rely on some auxiliary definitions given below. Once we have defined $\partial (u)$ for all $u\in {S}_{2}(X)$, we will then define $\partial (u)$ for all $u\in {C}_{2}(X)$ by the rule
$$\partial ({n}_{1}{u}_{1}+\mathrm{\cdots}+{n}_{r}{u}_{r})={n}_{1}\partial ({u}_{1})+\mathrm{\cdots}+{n}_{r}\partial ({u}_{r}),$$ |
just as we did for singular $1$-chains. This gives a homomorphism $\partial :{C}_{2}(X)\to {C}_{1}(X)$.
For a linear $3$-simplex $u=\u27e8{a}_{0},{a}_{1},{a}_{2},{a}_{3}\u27e9$, we will have
$$\partial (u)=\u27e8{a}_{1},{a}_{2},{a}_{3}\u27e9-\u27e8{a}_{0},{a}_{2},{a}_{3}\u27e9+\u27e8{a}_{0},{a}_{1},{a}_{3}\u27e9-\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9.$$ |
For a general linear $k$-simplex $u=\u27e8a\u27e9=\u27e8{a}_{0},\mathrm{\cdots},{a}_{k}\u27e9$, we will have
$$\partial (u)=\sum _{i=0}^{k}{(-1)}^{i}(\u27e8a\u27e9\text{with}{a}_{i}\text{omitted}).$$ |
For $0\le i\le n$ with $n>0$ we define ${\delta}_{i}:{\mathrm{\Delta}}_{n-1}\to {\mathrm{\Delta}}_{n}$ by
$${\delta}_{i}({t}_{0},\mathrm{\dots},{t}_{n-1})=({t}_{0},\mathrm{\dots},{t}_{i-1},0,{t}_{i},\mathrm{\dots},{t}_{n-1}).$$ |
Equivalently, we have
$$ |
Thus, the coordinates of ${\delta}_{i}(t)$ are the same as the coordinates of $t$, except that we insert a zero in position $i$.
In the case $n=1$ we have maps ${\delta}_{0},{\delta}_{1}:{\mathrm{\Delta}}_{0}=\{{e}_{0}\}=\{1\}\to {\mathrm{\Delta}}_{1}$. These are given by ${\delta}_{0}({e}_{0})={\delta}_{0}(1)=(0,1)={e}_{1}$ and ${\delta}_{1}({e}_{0})={\delta}_{1}(1)=(1,0)={e}_{0}$.
In the case $n=2$, we have
$${\delta}_{0}({t}_{0},{t}_{1})=(0,{t}_{0},{t}_{1})\mathit{\hspace{1em}\hspace{1em}}{\delta}_{1}({t}_{0},{t}_{1})=({t}_{0},0,{t}_{1})\mathit{\hspace{1em}\hspace{1em}}{\delta}_{2}({t}_{0},{t}_{1})=({t}_{0},{t}_{1},0).$$ |
Thus, the image of ${\delta}_{i}:{\mathrm{\Delta}}_{1}\to {\mathrm{\Delta}}_{2}$ is the edge of ${\mathrm{\Delta}}_{2}$ opposite the vertex ${e}_{i}$.
Similarly, in the case $n=3$, we have a map ${\delta}_{i}$ from the triangle ${\mathrm{\Delta}}_{2}$ to the tetrahedron ${\mathrm{\Delta}}_{3}$, and the image ${\delta}_{i}({\mathrm{\Delta}}_{2})$ is the face of the tetrahedron that is opposite the vertex ${\delta}_{i}$. The case $i=0$ is shown below.
Even more generally, we see that the map ${\delta}_{i}:{\mathrm{\Delta}}_{n-1}\to {\mathrm{\Delta}}_{n}$ gives a homeomorphism from ${\mathrm{\Delta}}_{n-1}$ to $\{t\in {\mathrm{\Delta}}_{n}|{t}_{i}=0\}$.
Consider an element $u\in {S}_{k}(X)$ (with $k>0$), or equivalently a continuous map $u:{\mathrm{\Delta}}_{k}\to X$. For each $i$ with $0\le i\le k$ we have a map ${\delta}_{i}:{\mathrm{\Delta}}_{k-1}\to {\mathrm{\Delta}}_{k}$ and we can compose this with $u$ to get a map $u\circ {\delta}_{i}:{\mathrm{\Delta}}_{k-1}\to X$, or in other words an element $u\circ {\delta}_{i}\in {S}_{k-1}(X)\subset {C}_{k-1}(X)$. We put
$$\partial (u)=\sum _{i=0}^{k}{(-1)}^{i}(u\circ {\delta}_{i})\in {C}_{k-1}(X).$$ |
More generally, given an element $u={\sum}_{p=1}^{r}{a}_{p}{u}_{p}\in {C}_{k}(X)$, we define $\partial (u)={\sum}_{p=1}^{r}{a}_{p}\partial ({u}_{p})\in {C}_{k-1}(X)$.
For a singular $1$-simplex $u:{\mathrm{\Delta}}_{1}\to X$ we have $\partial (u)=(u\circ {\delta}_{0})-(u\circ {\delta}_{1})$. Here ${\delta}_{0}$ sends the unique point of ${\mathrm{\Delta}}_{0}$ to ${e}_{1}$, so the map $u\circ {\delta}_{0}:{\mathrm{\Delta}}_{0}\to X$ corresponds to the point $u({e}_{1})\in X$. Similarly, ${\delta}_{1}$ sends the unique point of ${\mathrm{\Delta}}_{0}$ to ${e}_{0}$, so the map $u\circ {\delta}_{1}:{\mathrm{\Delta}}_{0}\to X$ corresponds to the point $u({e}_{0})\in X$. We therefore have $\partial (u)=u({e}_{1})-u({e}_{0})$, just as in Predefinition 10.11.
Now consider a linear $2$-simplex $u=\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9$, so
$$u({t}_{0},{t}_{1},{t}_{2})={t}_{0}{a}_{0}+{t}_{1}{a}_{1}+{t}_{2}{a}_{2}.$$ |
We find that
$(u\circ {\delta}_{0})({t}_{0},{t}_{1})$ | $=u(0,{t}_{0},{t}_{1})={t}_{0}{a}_{1}+{t}_{1}{a}_{2}=\u27e8{a}_{1},{a}_{2}\u27e9({t}_{0},{t}_{1})$ | ||
$(u\circ {\delta}_{1})({t}_{0},{t}_{1})$ | $=u({t}_{0},0,{t}_{1})={t}_{0}{a}_{0}+{t}_{1}{a}_{2}=\u27e8{a}_{0},{a}_{2}\u27e9({t}_{0},{t}_{1})$ | ||
$(u\circ {\delta}_{2})({t}_{0},{t}_{1})$ | $=u({t}_{0},{t}_{1},0)={t}_{0}{a}_{0}+{t}_{1}{a}_{1}=\u27e8{a}_{0},{a}_{1}\u27e9({t}_{0},{t}_{1}),$ |
so $u\circ {\delta}_{0}=\u27e8{a}_{1},{a}_{2}\u27e9$ and $u\circ {\delta}_{1}=\u27e8{a}_{0},{a}_{2}\u27e9$ and $u\circ {\delta}_{2}=\u27e8{a}_{0},{a}_{1}\u27e9$. This gives
$$\partial (u)=\u27e8{a}_{1},{a}_{2}\u27e9-\u27e8{a}_{0},{a}_{2}\u27e9+\u27e8{a}_{0},{a}_{1}\u27e9,$$ |
just as in Predefinition 10.11. It should be clear that the same pattern works for all $k$, giving
$$\partial (\u27e8{a}_{0},\mathrm{\dots},{a}_{k}\u27e9)=\sum _{i=0}^{k}{(-1)}^{i}(\u27e8{a}_{0},\mathrm{\dots},{a}_{k}\u27e9\text{with}{a}_{i}\text{omitted}).$$ |
The following result is crucial for the development of homology theory.
For all $u\mathrm{\in}{C}_{k}\mathit{}\mathrm{(}X\mathrm{)}$, we have ${\mathrm{\partial}}^{\mathrm{2}}\mathit{}\mathrm{(}u\mathrm{)}\mathrm{=}\mathrm{\partial}\mathit{}\mathrm{(}\mathrm{\partial}\mathit{}\mathrm{(}u\mathrm{)}\mathrm{)}\mathrm{=}\mathrm{0}$ in ${C}_{k\mathrm{-}\mathrm{2}}\mathit{}\mathrm{(}X\mathrm{)}$. Thus, the composite
$${C}_{k}(X)\stackrel{\partial}{\to}{C}_{k-1}(X)\stackrel{\partial}{\to}{C}_{k-2}(X)$$ |
is zero.
Recall that we defined ${C}_{j}(X)=0$ for $$, and any homomorphism to the zero group is automatically the zero homomorphism. Thus, the proposition has no content for $$. For the first nontrivial case, suppose that $X\subseteq {\mathbb{R}}^{N}$, and consider a linear $2$-simplex $u=\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9$. We then have
$\partial (u)$ | $=\u27e8{a}_{1},{a}_{2}\u27e9-\u27e8{a}_{0},{a}_{2}\u27e9+\u27e8{a}_{0},{a}_{1}\u27e9$ | ||
${\partial}^{2}(u)$ | $=\partial (\u27e8{a}_{1},{a}_{2}\u27e9)-\partial (\u27e8{a}_{0},{a}_{2}\u27e9)+\partial (\u27e8{a}_{0},{a}_{1}\u27e9)$ | ||
$=({a}_{2}-{a}_{1})-({a}_{2}-{a}_{0})+({a}_{1}-{a}_{0})=0.$ |
We will often use abbreviated notation for this kind of calculation, writing $012$ for $\u27e8{a}_{0},{a}_{1},{a}_{2}\u27e9$ and $02$ for $\u27e8{a}_{0},{a}_{2}\u27e9$, for example. With this notation, the above calculation becomes
$${\partial}^{2}(012)=\partial (12)-\partial (02)+\partial (01)=(2-1)-(2-0)+(1-0)=0.$$ |
We now discuss ${\partial}^{2}(u)$ where $u=\u27e8{a}_{0},{a}_{1},{a}_{2},{a}_{3},{a}_{4}\u27e9\in {C}_{4}(X)$, using the same kind of notation. First, we have
$$\partial (u)=1234-0234+0134-0124+0123.$$ |
We can write the terms of ${\partial}^{2}(u)$ in a square array, with $\partial (1234)$ in the first column, $\partial (-0234)$ in the second column, and so on. The result is as follows:
We find that the terms above the wavy line cancel in the indicated groups with the terms below the wavy line, leaving ${\partial}^{2}(u)=0$ as claimed.
If $$ then ${\delta}_{j}\mathit{}{\delta}_{i}\mathrm{=}{\delta}_{i}\mathit{}{\delta}_{j\mathrm{-}\mathrm{1}}\mathrm{:}{\mathrm{\Delta}}_{k\mathrm{-}\mathrm{2}}\mathrm{\to}{\mathrm{\Delta}}_{k}$.
Consider a point $t=({t}_{0},\mathrm{\dots},{t}_{k-2})\in {\mathrm{\Delta}}_{k-2}$. To form ${\delta}_{i}(t)$, we insert a zero in position $i$. To form ${\delta}_{j}({\delta}_{i}(t))$, we insert another zero in position $j$. Because $j>i$, inserting this second zero does not move the first zero, so we end up with zeros in positions $i$ and $j$.
Similarly, to form ${\delta}_{j-1}(t)$, we insert a zero in position $j-1$. To form ${\delta}_{i}({\delta}_{j-1}(t))$, we insert another zero in position $i$. As $j-1\ge i$ we see that the first zero is to the right of the point where we insert the second zero, so the first zero gets moved over by one space into position $j$. Thus, we again end up with zeros in positions $i$ and $j$. In the remaining positions, we have the numbers ${t}_{0},\mathrm{\dots},{t}_{k-2}$ in order. Thus, we have ${\delta}_{j}({\delta}_{i}(t))={\delta}_{i}({\delta}_{j-1}(t))$ as claimed. ∎
In the case where $(i,j,k)=(2,4,6)$ the claim is that ${\delta}_{4}{\delta}_{2}={\delta}_{2}{\delta}_{3}:{\mathrm{\Delta}}_{4}\to {\mathrm{\Delta}}_{6}$. Explicitly, for $t=({t}_{0},\mathrm{\dots},{t}_{4})\in {\mathrm{\Delta}}_{4}$ we have
We will now prove Proposition 10.16 in the case $k=4$. Consider a continuous map $u:{\mathrm{\Delta}}_{4}\to X$, or equivalently an element $u\in {S}_{4}(X)\subset {C}_{4}(X)$. We have
$$\partial (u)=u{\delta}_{0}-u{\delta}_{1}+u{\delta}_{2}-u{\delta}_{3}+u{\delta}_{4}.$$ |
We can write the terms of ${\partial}^{2}(u)$ in a square array, with $\partial (u{\delta}_{0})$ in the first column, $\partial (-u{\delta}_{1})$ in the second column, and so on. The result is as follows:
Lemma 10.18 gives us the following identities:
${\delta}_{1}{\delta}_{0}$ | $={\delta}_{0}{\delta}_{0}$ | ${\delta}_{2}{\delta}_{0}$ | $={\delta}_{0}{\delta}_{1}$ | ${\delta}_{3}{\delta}_{0}$ | $={\delta}_{0}{\delta}_{2}$ | ${\delta}_{4}{\delta}_{0}$ | $={\delta}_{0}{\delta}_{3}$ | ||
${\delta}_{2}{\delta}_{1}$ | $={\delta}_{1}{\delta}_{1}$ | ${\delta}_{3}{\delta}_{1}$ | $={\delta}_{1}{\delta}_{2}$ | ${\delta}_{4}{\delta}_{1}$ | $={\delta}_{1}{\delta}_{3}$ | ||||
${\delta}_{3}{\delta}_{2}$ | $={\delta}_{2}{\delta}_{2}$ | ${\delta}_{4}{\delta}_{2}$ | $={\delta}_{2}{\delta}_{3}$ | ||||||
${\delta}_{4}{\delta}_{3}$ | $={\delta}_{3}{\delta}_{3}$ |
Using this, we see that in the previous array, the terms above the wavy line cancel in the indicated groups with the terms below the wavy line, showing that ${\partial}^{2}(u)=0$ as claimed. This generalises the argument for linear simplices given in Example 10.17.
Consider a continuous map $u:{\mathrm{\Delta}}_{k}\to X$, or equivalently an element $u\in {S}_{k}(X)\subset {C}_{k}(X)$. We have
$${\partial}^{2}(u)=\sum _{j=0}^{k}{(-1)}^{j}\partial (u\circ {\delta}_{j})=\sum _{i=0}^{k-1}\sum _{j=0}^{k}{(-1)}^{i+j}u\circ {\delta}_{j}\circ {\delta}_{i}.$$ |
We can write this as $A+B$, where
$A$ | $$ | ||
$B$ | $={\displaystyle \sum _{0\le j\le i\le k-1}}{(-1)}^{i+j}u\circ {\delta}_{j}\circ {\delta}_{i}.$ |
Here $i$ and $j$ are just dummy variables, so we can rewrite $B$ as
$$B=\sum _{0\le q\le p\le k-1}{(-1)}^{p+q}u\circ {\delta}_{q}\circ {\delta}_{p}.$$ |
We now reindex again, taking $q=i$ and $p=j-1$. The condition $q\le p$ becomes $i\le j-1$ or equivalently $$. The condition $p\le k-1$ becomes $j-1\le k-1$ or equivalently $j\le k$. The sign ${(-1)}^{p+q}$ becomes ${(-1)}^{i+j-1}=-{(-1)}^{i+j}$. This gives
$$ |
However, Lemma 10.18 tells us that ${\delta}_{i}\circ {\delta}_{j-1}={\delta}_{j}\circ {\delta}_{i}$ here, so $B=-A$, so ${\partial}^{2}(u)=A+B=0$ as claimed.
This proves that ${\partial}^{2}(u)=0$ whenever $u$ is a singular $k$-simplex. More generally, and singular $k$-chain has the form $u={a}_{1}{u}_{1}+\mathrm{\cdots}+{a}_{r}{u}_{r}$ for some integers ${a}_{i}$ and singular $k$-simplices ${u}_{i}:{\mathrm{\Delta}}_{k}\to X$. We then have ${\partial}^{2}({u}_{i})=0$ for all $i$ and so ${\partial}^{2}(u)={\sum}_{i}{a}_{i}{\partial}^{2}({u}_{i})=0$. ∎
We say that an element $u\in {C}_{k}(X)$ is a $k$-cycle if $\partial (u)=0$. We write ${Z}_{k}(X)$ for the abelian group of $k$-cycles, so ${Z}_{k}(X)=\mathrm{ker}(\partial :{C}_{k}(X)\to {C}_{k-1}(X))$.
We say that an element $u\in {C}_{k}(X)$ is a $k$-boundary if there exists $v\in {C}_{k+1}(X)$ with $\partial (v)=u$. We write ${B}_{k}(X)$ for the abelian group of $k$-boundaries, so ${B}_{k}(X)=\mathrm{img}(\partial :{C}_{k+1}(X)\to {C}_{k}(X))$.
We note that if $u\in {B}_{k}(X)$ then $u=\partial (v)$ for some $v$, so $\partial (u)={\partial}^{2}(v)=0$ by Proposition 10.16, so $u\in {Z}_{k}(X)$. This means that ${B}_{k}(X)\le {Z}_{k}(X)$, so we can form the quotient abelian group ${H}_{k}(X)=({Z}_{k}(X))/({B}_{k}(X))$. We call this the $k$’th homology group of $X$.
The elements of ${H}_{k}(X)$ are cosets $z+{B}_{k}(X)$ with $z\in {Z}_{k}(X)$, so $z\in {C}_{k}(X)$ with $\partial (z)=0$. We will often write $[z]$ for $z+{B}_{k}(X)$. Before writing notation like $[z]$ one must check that $\partial (z)=0$; it is an error to use that notation in other cases. Note that $[z]=[{z}^{\prime}]$ iff $z-{z}^{\prime}\in {B}_{k}(X)$ iff there exists $w\in {C}_{k+1}(X)$ with $\partial (w)=z-{z}^{\prime}$.
There is essentially only one example that we can calculate directly from the definition.
If $X$ consists of a single point, then ${H}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}\mathbb{Z}$ and ${H}_{k}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}\mathrm{0}$ for $k\mathrm{\ne}\mathrm{0}$.
There is only one possible map from ${\mathrm{\Delta}}_{k}$ to $X$, sending all possible points in ${\mathrm{\Delta}}_{k}$ to the unique point of $X$. We call this map ${s}_{k}$, so ${S}_{k}(X)=\{{s}_{k}\}$ and ${C}_{k}(X)=\mathbb{Z}.{s}_{k}$ for all $k\ge 0$ (whereas ${C}_{k}(X)=0$ for $$ by definition). For $k>0$ we have $\partial ({s}_{k})={\sum}_{i=0}^{k}{(-1)}^{i}{s}_{k}\circ {\delta}_{i}$. Here ${s}_{k}\circ {\delta}_{i}$ is a map from ${\mathrm{\Delta}}_{k-1}$ to $X$ so it can only be equal to ${s}_{k-1}$. This gives
$\partial ({s}_{1})$ | $={s}_{0}-{s}_{0}=0$ | ||
$\partial ({s}_{2})$ | $={s}_{1}-{s}_{1}+{s}_{1}={s}_{1}$ | ||
$\partial ({s}_{3})$ | $={s}_{2}-{s}_{2}+{s}_{2}-{s}_{2}=0$ |
and so on. In general, we have $\partial ({s}_{2n+1})=0$ and $\partial ({s}_{2n+2})={s}_{2n+1}$. It follows that ${B}_{2n+1}(X)={Z}_{2n+1}(X)=\mathbb{Z}.{s}_{2n+1}$ and ${B}_{2n+2}(X)={Z}_{2n+2}(X)=0$. In particular, for all $k>0$ we have ${Z}_{k}(X)={B}_{k}(X)$ so the quotient group ${H}_{k}(X)=({Z}_{k}(X))/({B}_{k}(X))$ is trivial. On the other hand, ${Z}_{0}(X)=\mathbb{Z}.{s}_{0}$ and ${B}_{0}(X)=0$ so ${H}_{0}(X)=(\mathbb{Z}.{s}_{0})/0\simeq \mathbb{Z}$. All this can be tabulated as follows:
∎
We leave the following slight generalisation to the reader. Suppose that $X$ is a finite, discrete set of points, so that every continuous map ${\mathrm{\Delta}}_{k}\to X$ is constant. Then ${H}_{0}(X)={C}_{0}(X)=\mathbb{Z}\{X\}$, and ${H}_{k}(X)=0$ for $k\ne 0$.
We can also calculate ${H}_{0}(X)$ for all $X$.
There is a canonical isomorphism ${H}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\simeq}\mathbb{Z}\mathit{}\mathrm{\{}{\pi}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\}}$ for all topological spaces $X$. Thus, if $\mathrm{|}{\pi}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{|}\mathrm{=}r$ then ${H}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\simeq}{\mathbb{Z}}^{r}$.
This should not be a surprise. Both ${H}_{0}(X)$ and $\mathbb{Z}\{{\pi}_{0}(X)\}$ are ways of constructing an abelian group from $X$, in such a way that points connected by a path give the same element of the group. We just need to check that the technical differences between these two constructions do not affect the final answer.
First note that ${C}_{-1}(X)$ is zero by definition, so the map $\partial :{C}_{0}(X)\to {C}_{-1}(X)$ sends everything to zero, so ${Z}_{0}(X)={C}_{0}(X)$. This means that the quotient group ${H}_{0}(X)={Z}_{0}(X)/{B}_{0}(X)$ is the same as ${C}_{0}(X)/{B}_{0}(X)$.
Next, let $\pi :X\to {\pi}_{0}(X)$ be the usual quotient map, which sends every point $x\in X$ to the corresponding path component $[x]\in {\pi}_{0}(X)$. We can extend this linearly to give a homomorphism $\pi :\mathbb{Z}\{X\}=\mathbb{Z}\{{S}_{0}(X)\}={C}_{0}(X)\to \mathbb{Z}\{{\pi}_{0}(X)\}$, by the rule
$$\pi ({n}_{1}{x}_{1}+\mathrm{\cdots}+{n}_{p}{x}_{p})={n}_{1}\pi ({x}_{1})+\mathrm{\cdots}+{n}_{p}\pi ({x}_{p})={n}_{1}[{x}_{1}]+\mathrm{\cdots}+{n}_{p}[{x}_{p}]\in \mathbb{Z}\{{\pi}_{0}(X)\}.$$ |
We will show that $\pi $ is surjective, with kernel ${B}_{0}(X)$. Assuming this, the First Isomorphism Theorem will give us an isomorphism from ${C}_{0}(X)/{B}_{0}(X)={H}_{0}(X)$ to $\mathbb{Z}\{{\pi}_{0}(X)\}$, as required.
Next, for each path component $c\in {\pi}_{0}(X)$, we choose a point $\sigma (c)\in c$, so $c=[\sigma (c)]$. This means that the composite
$${\pi}_{0}(X)\stackrel{\mathit{\sigma}}{\to}X\stackrel{\mathit{\pi}}{\to}{\pi}_{0}(X)$$ |
is the identity. We can also extend $\sigma $ linearly to give a homomorphism $\sigma :\mathbb{Z}\{{\pi}_{0}(X)\}\to {C}_{0}(X)$ by the rule $\sigma ({n}_{1}{c}_{1}+\mathrm{\cdots}+n+p{c}_{p})={n}_{1}\sigma ({c}_{1})+\mathrm{\cdots}+{n}_{p}\sigma ({c}_{p})$. In this context, we see that the composite
$$\mathbb{Z}\{{\pi}_{0}(X)\}\stackrel{\mathit{\sigma}}{\to}{C}_{0}(X)\stackrel{\mathit{\pi}}{\to}\mathbb{Z}\{{\pi}_{0}(X)\}$$ |
is again the identity. In particular, any element $u\in \mathbb{Z}\{{\pi}_{0}(X)\}$ is the same as $\pi (\sigma (u))$, so it is in the image of $\pi $; this proves that $\pi $ is surjective.
Now suppose we have a path $v\in {S}_{1}(X)$. We then have $\partial (v)=v({e}_{1})-v({e}_{0})\in {C}_{0}(X)$, so $\pi (\partial (v))=\pi (v({e}_{1}))-\pi (v({e}_{0}))=[v({e}_{1})]-[v({e}_{0})]\in \mathbb{Z}\{{\pi}_{0}(X)\}$. However, we have a path $v$ joining $v({e}_{0})$ to $v({e}_{1})$, so the corresponding path components are the same, so $\pi (\partial (v))=0$. As everything is extended linearly, the rule $\pi (\partial (v))=0$ remains valid for all $v\in {C}_{1}(X)$. The image of $\partial :{C}_{1}(X)\to {C}_{0}(X)$ is ${B}_{0}(X)$, so this means that $\pi ({B}_{0}(X))=0$, or equivalently ${B}_{0}(X)\le \mathrm{ker}(\pi )$.
Next, consider a point $x\in X$ and the corresponding path component $c=[x]=\pi (x)$. The points $x$ and $\sigma (c)=\sigma (\pi (x))$ both lie in the same path component $c$, so there must exist a path from $\sigma (\pi (x))$ to $x$ in $X$. We choose such a path and call it $\gamma (x)$. This defines a function $\gamma $ from $X$ to the set ${S}_{1}(X)$ of paths in $X$, which we extend linearly to get a hoomorphism $\gamma :{C}_{0}(X)\to {C}_{1}(X)$. For any point $x$ we know that $\gamma (x)$ runs from $\sigma (\pi (x))$ to $x$, so $\partial \gamma (x)=x-\sigma (\pi (x))$. As everything is extended linearly, the rule $\partial (\gamma (u))=u-\sigma (\pi (u))$ is valid for all $u\in {C}_{0}(X)$. In particular, if $u\in \mathrm{ker}(\pi )$ then $\pi (u)=0$ so this simplifies to $\partial (\gamma (u))=u$, proving that $u$ is in the image of $\partial $, or in other words $u\in {B}_{0}(X)$.
We can now conclude that $\pi $ is surjective with kernel ${B}_{0}(X)$. By the First Isomorphism Theorem, there is a well-defined homomorphism $\overline{\pi}:{H}_{0}(X)={C}_{0}(X)/{B}_{0}(X)\to \mathbb{Z}\{{\pi}_{0}(X)\}$ given by $\overline{\pi}(u+{B}_{0}(X))=\pi (u)$ for all $u\in {C}_{0}(X)$, and this is in fact an isomorphism. ∎
The above proof can be illustrated by the following diagram. It shows a space $X$ with three path components $A$, $B$ and $C$, so ${\pi}_{0}(X)=\{A,B,C\}$ and
$$\mathbb{Z}\{{\pi}_{0}(X)\}=\{kA+nB+mC|k,n,m\in \mathbb{Z}\}\simeq {\mathbb{Z}}^{3}.$$ |
We have chosen points $\sigma (A)\in A$ and $\sigma (B)\in B$ and $\sigma (C)\in C$, so $A=[\sigma (A)]$ and $B=[\sigma (B)]$ and $C=[\sigma (C)]$. To say the same thing in different notation, we have $\pi (\sigma (A))=A$ and $\pi (\sigma (B))=B$ and $\pi (\sigma (C))=C$, so $\pi \circ \sigma =\mathrm{id}$. The points ${a}_{1}$ and ${a}_{2}$ also lie in $A$, so $[{a}_{1}]=[{a}_{2}]=A$, or equivalently $\pi ({a}_{1})=\pi ({a}_{2})=A$. The path $\gamma ({a}_{1})$ runs from $\sigma (A)=\sigma (\pi ({a}_{1}))$ to ${a}_{1}$. Similarly, we have $\pi ({b}_{1})=\pi ({b}_{2})=\pi ({b}_{3})=B$, and we have labelled a path $\gamma ({b}_{3})$ running from $\sigma (\pi ({b}_{3}))=\sigma (B)$ to ${b}_{3}$.
A typical example of an element of $\mathrm{ker}(\pi :{C}_{0}(X)\to \mathbb{Z}\{{\pi}_{0}(X)\})$ could be the element $u={a}_{1}-{a}_{2}+{b}_{1}+{b}_{3}-2{b}_{2}$. This has $\gamma (u)=\gamma ({a}_{1})-\gamma ({a}_{2})+\gamma ({b}_{1})+\gamma ({b}_{3})-2\gamma ({b}_{2})$, so
$\partial (\gamma (u))$ | $=({a}_{1}-\sigma (A))-({a}_{2}-\sigma (A))+({b}_{1}-\sigma (B))+({b}_{3}-\sigma (B))-2({b}_{2}-\sigma (B))$ | ||
$={a}_{1}-{a}_{2}+{b}_{1}+{b}_{3}-2{b}_{2}=u,$ |
so $u=\partial (\gamma (u))\in \mathrm{img}(\partial )={B}_{0}(X)$. This illustrates the fact that $\mathrm{ker}(\pi )=\mathrm{img}(\partial )$, which is a key step in our proof of Proposition 10.25.
We next discuss homology classes of paths, revisiting Remark 10.10.
Let $X$ be a topological space.
For any $a\in X$ the constant path ${c}_{a}\in {S}_{1}(X)\subseteq {C}_{1}(X)$ actually lies in ${B}_{1}(X)$, so ${c}_{a}+{B}_{1}(X)=0$ in the quotient group ${C}_{1}(X)/{B}_{1}(X)$.
For any path $u:a\rightsquigarrow b$ in $X$ with reversed path $\overline{u}:b\rightsquigarrow a$, we have $u+\overline{u}\in {B}_{1}(X)$ so $\overline{u}+{B}_{1}(X)=-u+{B}_{1}(X)$ in ${C}_{1}(X)/{B}_{1}(X)$.
For any paths $u:a\rightsquigarrow b$ and $v:b\rightsquigarrow c$ we have $(u*v)+{B}_{1}(X)=(u+{B}_{1}(X))+(v+{B}_{1}(X))$ in ${C}_{1}(X)/{B}_{1}(X)$.
Video (Path homotopy, loop homotopy and homology)
Let $X$ be a topological space. A loop in $X$ is a path $u:{\mathrm{\Delta}}_{1}\to X$ with $u({e}_{0})=u({e}_{1})$, so that $\partial (u)=0$, so we have a coset $[u]=u+{B}_{1}(X)\in {H}_{1}(X)$. If $u({e}_{0})=u({e}_{1})=a$, we say that $u$ is a loop based at $a$.
Let $X$ be a path connected space, and let $a$ be a point in $X$. Then for every $h\mathrm{\in}{H}_{\mathrm{1}}\mathit{}\mathrm{(}X\mathrm{)}$ there exists a loop $u$ based at $a$ with $h\mathrm{=}\mathrm{[}u\mathrm{]}$. Moreover, if $u$ and $v$ are loops based at $a$ then so are ${c}_{a}$, $\overline{u}$ and $u\mathrm{*}v$, and we have $\mathrm{[}{c}_{a}\mathrm{]}\mathrm{=}\mathrm{0}$ and $\mathrm{[}\overline{u}\mathrm{]}\mathrm{=}\mathrm{-}\mathrm{[}u\mathrm{]}$ and $\mathrm{[}u\mathrm{*}v\mathrm{]}\mathrm{=}\mathrm{[}u\mathrm{]}\mathrm{+}\mathrm{[}v\mathrm{]}$ in ${H}_{\mathrm{1}}\mathit{}\mathrm{(}X\mathrm{)}$.
Let $L$ be the subset of ${H}_{1}(X)$ consisting of classes that can be expressed as $[u]$ for some loop $u$ based at $a$. We must show that this is all of ${H}_{1}(X)$.
It is clear that if $u$ and $v$ are loops based at $a$, then so are ${c}_{a}$, $\overline{u}$ and $u*v$. By specialising Lemma 10.27, we see that $[{c}_{a}]=0$ and $[\overline{u}]=-[u]$ and $[u*v]=[u]+[v]$ in ${H}_{1}(X)$. It follows from this that $L$ is a subgroup of ${H}_{1}(X)$.
Now let $v$ be a loop based at a point $b\in X$ which may be different from $a$. As $X$ is path connected, we can choose a path $m$ from $a$ to $b$. The path $u=(m*v)*\overline{m}$ is then a loop based at $a$, and using Lemma 10.27 again we see that
$$u+{B}_{1}(X)=m+v-m+{B}_{1}(X)=v+{B}_{1}(X),$$ |
or in other word $[u]=[v]$ in ${H}_{1}(X)$. This proves that $L$ contains all loops, irrespective of the base point.
Now let $h$ be an arbitrary element of ${H}_{1}(X)$. We can write $h$ as $z+{B}_{1}(X)$, where $z$ is a $\mathbb{Z}$-linear combination of paths in $X$. Any term with negative coefficient like $-m.u$ can be replaced by $+m.\overline{u}$ without affecting the coset, so we can assume that all coefficients are positive. Then we can replace any term like $m.u$ by $u$ repeated $m$ times; this gives an expression like
$$h={u}_{1}+\mathrm{\cdots}+{u}_{n}+{B}_{1}(X)$$ |
for some list of paths ${u}_{i}$. As this is a homology class, the representing chain must be a cycle, so we must have $\partial ({u}_{1}+\mathrm{\cdots}+{u}_{n})=0$ in ${C}_{0}(X)$. As $\partial ({u}_{i})={u}_{i}({e}_{1})-{u}_{i}({e}_{0})$, this means that
$${u}_{1}({e}_{1})+\mathrm{\cdots}+{u}_{n}({e}_{1})={u}_{1}({e}_{0})+\mathrm{\cdots}+{u}_{n}({e}_{0}).$$ |
As this is happening in the free abelian group $\mathbb{Z}\{X\}$, the terms on the left hand side must just be a permutation of those on the right hand side, so we have a permutation $\sigma $ of $\{1,\mathrm{\dots},n\}$ with ${u}_{i}({e}_{1})={u}_{\sigma (i)}({e}_{0})$ for all $i$. We can now write $\sigma $ as a product of disjoint cycles. If one of these cycles is $(ijkl)$, for example, then the paths ${u}_{i}$, ${u}_{j}$, ${u}_{k}$ and ${u}_{l}$ meet end-to-end and so can be joined together to form a loop $(({u}_{i}*{u}_{j})*{u}_{k})*{u}_{l}$ which is congruent to ${u}_{i}+{u}_{j}+{u}_{k}+{u}_{l}$ modulo ${B}_{1}(X)$. By doing this for all cycles, we see that $h$ can be expressed as a sum of loops (probably with different basepoints). Our earlier discussion shows that each of these loops lies in $L$ and then that the sum lies in $L$, so $h\in L$ as claimed. ∎
Let $u:{\mathrm{\Delta}}_{1}\to X$ be a loop based at $a$. A filling in of $u$ is a map $v:{\mathrm{\Delta}}_{2}\to X$ with $v\circ {\delta}_{0}=u$ and $v\circ {\delta}_{1}=v\circ {\delta}_{2}={c}_{a}$.
If $u$ can be filled in, then $\mathrm{[}u\mathrm{]}\mathrm{=}\mathrm{0}$ in ${H}_{\mathrm{1}}\mathit{}\mathrm{(}X\mathrm{)}$.
Let $v$ be a filling in of $u$. Then
$$\partial (v)=v\circ {\delta}_{0}-v\circ {\delta}_{1}+v\circ {\delta}_{2}=u-{c}_{a}+{c}_{a}=u,$$ |
so $u\in {B}_{1}(X)$, so $[u]=u+{B}_{1}(X)=0$. ∎