# 10. Homology

Video (Definition 10.1 to Example 10.9)

###### Definition 10.1.

Let $X$ be a space. A singular $k$-simplex in $X$ is a continuous map $u\colon\Delta_{k}\to X$. We write $S_{k}(X)$ for the set of singular $k$-simplices in $X$.

###### Example 10.2.

Recall that $\Delta_{0}$ is the set $\{e_{0}\}$ with just one point. To give a function $u\colon\Delta_{0}\to X$ is the same as to give a point $u(e_{0})\in X$; this, we can identify $S_{0}(X)$ with $X$.

###### Example 10.3.

As usual, we identify $\Delta_{1}$ with $[0,1]$, with the point $(1-t,t)\in\Delta_{1}$ corresponding to the point $t\in[0,1]$. Thus, a singular $1$-simplex in $X$ is the same as a continuous map $u\colon[0,1]\to X$, or in other words a path in $X$. This means that $S_{1}(X)$ is the set of all possible paths in $X$.

###### Example 10.4.

Suppose that $a_{0},\dotsc,a_{k}\in{\mathbb{R}}^{N}$. We can then define a map

 $\langle a\rangle=\langle a_{0},\dotsc,a_{k}\rangle\colon\Delta_{k}\to{\mathbb{% R}}^{N}$

(or in other words an element $\langle a\rangle\in S_{k}{\mathbb{R}}^{N}$) by

 $\langle a\rangle(t_{0},\dotsc,t_{n})=t_{0}a_{0}+\dotsb+t_{k}a_{k}.$

We call maps of this type linear simplices.

In the case $k=0$ we have $S_{0}(X)=X$ and the map $\langle a_{0}\rangle$ just corresponds to the point $a_{0}$. In the case $k=1$, the map $\langle a_{0},a_{1}\rangle$ corresponds to the straight line path from $a_{0}$ to $a_{1}$. In the case $k=2$, the image of the map $\langle a_{0},a_{1},a_{2}\rangle\colon\Delta_{2}\to{\mathbb{R}}^{N}$ is the triangle with vertices $a_{0}$, $a_{1}$ and $a_{2}$.

Now suppose that $X\subseteq{\mathbb{R}}^{N}$ and that $a_{0},\dotsc,a_{k}\in X$. It may or may not happen that the image of the map $\langle a\rangle\colon\Delta_{k}\to{\mathbb{R}}^{N}$ actually lies in $X$; this must be checked carefully in any context where we want to use this construction. If so, we can regard $\langle a\rangle$ as an element of $S_{k}(X)$.

###### Example 10.5.

This picture shows a space $X\subset{\mathbb{R}}^{2}$, together with:

• A linear $2$-simplex $\langle a_{0},a_{1},a_{2}\rangle\in S_{2}{\mathbb{R}}^{2}$, which is not an element of $S_{2}(X)$.

• Another linear $2$-simplex $\langle b_{0},b_{1},b_{2}\rangle\in S_{2}(X)\subset S_{2}({\mathbb{R}}^{2})$.

• A nonlinear $1$-simplex $u\in S_{1}(X)$.

###### Definition 10.6.

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.)

###### Remark 10.7.

Suppose that $P$ is finite, say $P=\{p_{1}\dotsc,p_{n}\}$. We then have an isomorphism $\phi\colon{\mathbb{Z}}^{n}\to{\mathbb{Z}}\{P\}$ given by

 $\phi(a_{1},\dotsc,a_{n})=a_{1}p_{1}+\dotsb+a_{n}p_{n}.$

However, we will most often be considering cases where $P$ is infinite.

###### Definition 10.8.

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 $k<0$.

###### Example 10.9.

Consider again the picture in Example 10.5:

• The expression $6a_{1}-4b_{2}+7c_{1}\in C_{0}(X)$ is a singular $0$-chain.

• The expression $3\langle a_{0},a_{2}\rangle-\langle b_{0},b_{1}\rangle+u\in C_{1}(X)$ is a singular $1$-chain.

• The expression $\langle b_{0},b_{1},b_{2}\rangle\in S_{2}(X)\subset C_{2}(X)$ is a singular $2$-chain.

• No expression involving $\langle a_{0},a_{1}\rangle$ gives a singular chain in $X$, because the straight line from $a_{0}$ to $a_{1}$ is not contained in $X$.

###### Remark 10.10.

Suppose we have paths $u\colon a\rightsquigarrow b$ and $v\colon b\rightsquigarrow c$ in $X$. We can reverse $u$ to get a path $\overline{u}\colon b\rightsquigarrow a$, or we can join $u$ and $v$ to get a path $u*v\colon a\rightsquigarrow c$.

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.

Video (Predefinition 10.11 to Example 10.15)

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 $k=2$.

###### Predefinition 10.11.
• For $u\in C_{0}(X)={\mathbb{Z}}\{X\}$ we just define $\partial u=0$.

• Now consider a singular $1$-simplex $u\colon\Delta_{1}\to X$. 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 $u=n_{1}u_{1}+\dotsb+n_{r}u_{r}$, with $u_{i}\colon\Delta_{1}\to X$ and $n_{i}\in{\mathbb{Z}}$. We then put

 $\partial(u)=n_{1}\partial(u_{1})+\dotsb+n_{r}\partial(u_{r})=\sum_{i=1}^{r}n_{% i}(u_{i}(e_{1})-u_{i}(e_{0})).$

This defines a homomorphism $\partial\colon C_{1}(X)\to C_{0}(X)$.

Note that for a linear $1$-simplex $\langle a_{0},a_{1}\rangle$, we just have $\partial(\langle a_{0},a_{1}\rangle)=a_{1}-a_{0}$. Thus, in the picture below we have a $1$-chain

 $u=\langle a_{0},a_{1}\rangle+\langle a_{1},a_{2}\rangle+\langle a_{2},a_{3}% \rangle+\langle a_{3},a_{4}\rangle+\langle a_{4},a_{5}\rangle+\langle a_{5},a_% {6}\rangle$

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 $u=\langle a_{0},a_{1},a_{2}\rangle$ is a linear $2$-simplex. In this case, we define

 $\partial(\langle a_{0},a_{1},a_{2}\rangle)=\langle a_{1},a_{2}\rangle-\langle a% _{0},a_{2}\rangle+\langle a_{0},a_{1}\rangle.$

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

 $\partial(n_{1}u_{1}+\dotsb+n_{r}u_{r})=n_{1}\partial(u_{1})+\dotsb+n_{r}% \partial(u_{r}),$

just as we did for singular $1$-chains. This gives a homomorphism $\partial\colon C_{2}(X)\to C_{1}(X)$.

• For a linear $3$-simplex $u=\langle a_{0},a_{1},a_{2},a_{3}\rangle$, we will have

 $\partial(u)=\langle a_{1},a_{2},a_{3}\rangle-\langle a_{0},a_{2},a_{3}\rangle+% \langle a_{0},a_{1},a_{3}\rangle-\langle a_{0},a_{1},a_{2}\rangle.$

For a general linear $k$-simplex $u=\langle a\rangle=\langle a_{0},\dotsb,a_{k}\rangle$, we will have

 $\partial(u)=\sum_{i=0}^{k}(-1)^{i}(\langle a\rangle\text{ with }a_{i}\text{ % omitted }).$
###### Definition 10.12.

For $0\leq i\leq n$ with $n>0$ we define $\delta_{i}\colon\Delta_{n-1}\to\Delta_{n}$ by

 $\delta_{i}(t_{0},\dotsc,t_{n-1})=(t_{0},\dotsc,t_{i-1},0,t_{i},\dotsc,t_{n-1}).$

Equivalently, we have

 $\delta_{i}(t)_{j}=\begin{cases}t_{j}&\text{ if }ji.\end{cases}$

Thus, the coordinates of $\delta_{i}(t)$ are the same as the coordinates of $t$, except that we insert a zero in position $i$.

###### Example 10.13.

In the case $n=1$ we have maps $\delta_{0},\delta_{1}\colon\Delta_{0}=\{e_{0}\}=\{1\}\to\Delta_{1}$. 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

 $\delta_{0}(t_{0},t_{1})=(0,t_{0},t_{1})\qquad\delta_{1}(t_{0},t_{1})=(t_{0},0,% t_{1})\qquad\delta_{2}(t_{0},t_{1})=(t_{0},t_{1},0).$

Thus, the image of $\delta_{i}\colon\Delta_{1}\to\Delta_{2}$ is the edge of $\Delta_{2}$ opposite the vertex $e_{i}$.

Similarly, in the case $n=3$, we have a map $\delta_{i}$ from the triangle $\Delta_{2}$ to the tetrahedron $\Delta_{3}$, and the image $\delta_{i}(\Delta_{2})$ 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 $\delta_{i}\colon\Delta_{n-1}\to\Delta_{n}$ gives a homeomorphism from $\Delta_{n-1}$ to $\{t\in\Delta_{n}\;|\;t_{i}=0\}$.

###### Definition 10.14.

Consider an element $u\in S_{k}(X)$ (with $k>0$), or equivalently a continuous map $u\colon\Delta_{k}\to X$. For each $i$ with $0\leq i\leq k$ we have a map $\delta_{i}\colon\Delta_{k-1}\to\Delta_{k}$ and we can compose this with $u$ to get a map $u\circ\delta_{i}\colon\Delta_{k-1}\to X$, 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 $\partial(u)=\sum_{p=1}^{r}a_{p}\partial(u_{p})\in C_{k-1}(X)$.

###### Example 10.15.
• For a singular $1$-simplex $u\colon\Delta_{1}\to X$ we have $\partial(u)=(u\circ\delta_{0})-(u\circ\delta_{1})$. Here $\delta_{0}$ sends the unique point of $\Delta_{0}$ to $e_{1}$, so the map $u\circ\delta_{0}\colon\Delta_{0}\to X$ corresponds to the point $u(e_{1})\in X$. Similarly, $\delta_{1}$ sends the unique point of $\Delta_{0}$ to $e_{0}$, so the map $u\circ\delta_{1}\colon\Delta_{0}\to X$ 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 $u=\langle a_{0},a_{1},a_{2}\rangle$, so

 $u(t_{0},t_{1},t_{2})=t_{0}a_{0}+t_{1}a_{1}+t_{2}a_{2}.$

We find that

 $\displaystyle(u\circ\delta_{0})(t_{0},t_{1})$ $\displaystyle=u(0,t_{0},t_{1})=t_{0}a_{1}+t_{1}a_{2}=\langle a_{1},a_{2}% \rangle(t_{0},t_{1})$ $\displaystyle(u\circ\delta_{1})(t_{0},t_{1})$ $\displaystyle=u(t_{0},0,t_{1})=t_{0}a_{0}+t_{1}a_{2}=\langle a_{0},a_{2}% \rangle(t_{0},t_{1})$ $\displaystyle(u\circ\delta_{2})(t_{0},t_{1})$ $\displaystyle=u(t_{0},t_{1},0)=t_{0}a_{0}+t_{1}a_{1}=\langle a_{0},a_{1}% \rangle(t_{0},t_{1}),$

so $u\circ\delta_{0}=\langle a_{1},a_{2}\rangle$ and $u\circ\delta_{1}=\langle a_{0},a_{2}\rangle$ and $u\circ\delta_{2}=\langle a_{0},a_{1}\rangle$. This gives

 $\partial(u)=\langle a_{1},a_{2}\rangle-\langle a_{0},a_{2}\rangle+\langle a_{0% },a_{1}\rangle,$

just as in Predefinition 10.11. It should be clear that the same pattern works for all $k$, giving

 $\partial(\langle a_{0},\dotsc,a_{k}\rangle)=\sum_{i=0}^{k}(-1)^{i}(\langle a_{% 0},\dotsc,a_{k}\rangle\text{ with }a_{i}\text{ omitted}).$

The following result is crucial for the development of homology theory.

Video (Proposition 10.16 to Definition 10.21)

###### Proposition 10.16.

For all $u\in C_{k}(X)$, we have $\partial^{2}(u)=\partial(\partial(u))=0$ in $C_{k-2}(X)$. Thus, the composite

 $C_{k}(X)\xrightarrow{\partial}C_{k-1}(X)\xrightarrow{\partial}C_{k-2}(X)$

is zero.

###### Example 10.17.

Recall that we defined $C_{j}(X)=0$ for $j<0$, and any homomorphism to the zero group is automatically the zero homomorphism. Thus, the proposition has no content for $k<2$. For the first nontrivial case, suppose that $X\subseteq{\mathbb{R}}^{N}$, and consider a linear $2$-simplex $u=\langle a_{0},a_{1},a_{2}\rangle$. We then have

 $\displaystyle\partial(u)$ $\displaystyle=\langle a_{1},a_{2}\rangle-\langle a_{0},a_{2}\rangle+\langle a_% {0},a_{1}\rangle$ $\displaystyle\partial^{2}(u)$ $\displaystyle=\partial(\langle a_{1},a_{2}\rangle)-\partial(\langle a_{0},a_{2% }\rangle)+\partial(\langle a_{0},a_{1}\rangle)$ $\displaystyle=(a_{2}-a_{1})-(a_{2}-a_{0})+(a_{1}-a_{0})=0.$

We will often use abbreviated notation for this kind of calculation, writing $012$ for $\langle a_{0},a_{1},a_{2}\rangle$ and $02$ for $\langle a_{0},a_{2}\rangle$, 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 $u=\langle a_{0},a_{1},a_{2},a_{3},a_{4}\rangle\in C_{4}(X)$, 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.

###### Lemma 10.18.

If $0\leq i then $\delta_{j}\delta_{i}=\delta_{i}\delta_{j-1}\colon\Delta_{k-2}\to\Delta_{k}$.

###### Proof.

Consider a point $t=(t_{0},\dotsc,t_{k-2})\in\Delta_{k-2}$. 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 $j-1\geq i$ 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 $t_{0},\dotsc,t_{k-2}$ in order. Thus, we have $\delta_{j}(\delta_{i}(t))=\delta_{i}(\delta_{j-1}(t))$ as claimed. ∎

###### Example 10.19.

In the case where $(i,j,k)=(2,4,6)$ the claim is that $\delta_{4}\delta_{2}=\delta_{2}\delta_{3}\colon\Delta_{4}\to\Delta_{6}$. Explicitly, for $t=(t_{0},\dotsc,t_{4})\in\Delta_{4}$ we have

###### Example 10.20.

We will now prove Proposition 10.16 in the case $k=4$. Consider a continuous map $u\colon\Delta_{4}\to X$, 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:

 $\displaystyle\delta_{1}\delta_{0}$ $\displaystyle=\delta_{0}\delta_{0}$ $\displaystyle\delta_{2}\delta_{0}$ $\displaystyle=\delta_{0}\delta_{1}$ $\displaystyle\delta_{3}\delta_{0}$ $\displaystyle=\delta_{0}\delta_{2}$ $\displaystyle\delta_{4}\delta_{0}$ $\displaystyle=\delta_{0}\delta_{3}$ $\displaystyle\delta_{2}\delta_{1}$ $\displaystyle=\delta_{1}\delta_{1}$ $\displaystyle\delta_{3}\delta_{1}$ $\displaystyle=\delta_{1}\delta_{2}$ $\displaystyle\delta_{4}\delta_{1}$ $\displaystyle=\delta_{1}\delta_{3}$ $\displaystyle\delta_{3}\delta_{2}$ $\displaystyle=\delta_{2}\delta_{2}$ $\displaystyle\delta_{4}\delta_{2}$ $\displaystyle=\delta_{2}\delta_{3}$ $\displaystyle\delta_{4}\delta_{3}$ $\displaystyle=\delta_{3}\delta_{3}$

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.

###### Proof of Proposition 10.16.

Consider a continuous map $u\colon\Delta_{k}\to X$, 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

 $\displaystyle A$ $\displaystyle=\sum_{0\leq i $\displaystyle B$ $\displaystyle=\sum_{0\leq j\leq i\leq k-1}(-1)^{i+j}u\circ\delta_{j}\circ% \delta_{i}.$

Here $i$ and $j$ are just dummy variables, so we can rewrite $B$ as

 $B=\sum_{0\leq q\leq p\leq k-1}(-1)^{p+q}u\circ\delta_{q}\circ\delta_{p}.$

We now reindex again, taking $q=i$ and $p=j-1$. The condition $q\leq p$ becomes $i\leq j-1$ or equivalently $i. The condition $p\leq k-1$ becomes $j-1\leq k-1$ or equivalently $j\leq k$. The sign $(-1)^{p+q}$ becomes $(-1)^{i+j-1}=-(-1)^{i+j}$. This gives

 $B=-\sum_{0\leq i

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 $u=a_{1}u_{1}+\dotsb+a_{r}u_{r}$ for some integers $a_{i}$ and singular $k$-simplices $u_{i}\colon\Delta_{k}\to X$. We then have $\partial^{2}(u_{i})=0$ for all $i$ and so $\partial^{2}(u)=\sum_{i}a_{i}\partial^{2}(u_{i})=0$. ∎

###### Definition 10.21.
• (a)

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 $Z_{k}(X)=\ker(\partial\colon C_{k}(X)\to C_{k-1}(X))$.

• (b)

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 $B_{k}(X)=\operatorname{img}(\partial\colon C_{k+1}(X)\to C_{k}(X))$.

• (c)

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 $B_{k}(X)\leq Z_{k}(X)$, 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 $X$.

###### Remark 10.22.

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.

###### Proposition 10.23.

If $X$ consists of a single point, then $H_{0}(X)={\mathbb{Z}}$ and $H_{k}(X)=0$ for $k\neq 0$.

###### Proof.

There is only one possible map from $\Delta_{k}$ to $X$, sending all possible points in $\Delta_{k}$ 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 $k\geq 0$ (whereas $C_{k}(X)=0$ for $k<0$ 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 $\Delta_{k-1}$ to $X$ so it can only be equal to $s_{k-1}$. This gives

 $\displaystyle\partial(s_{1})$ $\displaystyle=s_{0}-s_{0}=0$ $\displaystyle\partial(s_{2})$ $\displaystyle=s_{1}-s_{1}+s_{1}=s_{1}$ $\displaystyle\partial(s_{3})$ $\displaystyle=s_{2}-s_{2}+s_{2}-s_{2}=0$

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:

###### Remark 10.24.

We leave the following slight generalisation to the reader. Suppose that $X$ is a finite, discrete set of points, so that every continuous map $\Delta_{k}\to X$ is constant. Then $H_{0}(X)=C_{0}(X)={\mathbb{Z}}\{X\}$, and $H_{k}(X)=0$ for $k\neq 0$.

We can also calculate $H_{0}(X)$ for all $X$.

###### Proposition 10.25.

There is a canonical isomorphism $H_{0}(X)\simeq{\mathbb{Z}}\{\pi_{0}(X)\}$ for all topological spaces $X$. Thus, if $|\pi_{0}(X)|=r$ then $H_{0}(X)\simeq{\mathbb{Z}}^{r}$.

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.

###### Proof.

First note that $C_{-1}(X)$ is zero by definition, so the map $\partial\colon C_{0}(X)\to C_{-1}(X)$ 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 $\pi\colon X\to\pi_{0}(X)$ 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 $\pi\colon{\mathbb{Z}}\{X\}={\mathbb{Z}}\{S_{0}(X)\}=C_{0}(X)\to{\mathbb{Z}}\{% \pi_{0}(X)\}$, by the rule

 $\pi(n_{1}x_{1}+\dotsb+n_{p}x_{p})=n_{1}\pi(x_{1})+\dotsb+n_{p}\pi(x_{p})=n_{1}% [x_{1}]+\dotsb+n_{p}[x_{p}]\in{\mathbb{Z}}\{\pi_{0}(X)\}.$

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

 $\pi_{0}(X)\xrightarrow{\sigma}X\xrightarrow{\pi}\pi_{0}(X)$

is the identity. We can also extend $\sigma$ linearly to give a homomorphism $\sigma\colon{\mathbb{Z}}\{\pi_{0}(X)\}\to C_{0}(X)$ by the rule $\sigma(n_{1}c_{1}+\dotsb+n+pc_{p})=n_{1}\sigma(c_{1})+\dotsb+n_{p}\sigma(c_{p})$. In this context, we see that the composite

 ${\mathbb{Z}}\{\pi_{0}(X)\}\xrightarrow{\sigma}C_{0}(X)\xrightarrow{\pi}{% \mathbb{Z}}\{\pi_{0}(X)\}$

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 $\partial\colon C_{1}(X)\to C_{0}(X)$ is $B_{0}(X)$, so this means that $\pi(B_{0}(X))=0$, or equivalently $B_{0}(X)\leq\ker(\pi)$.

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 $\gamma\colon C_{0}(X)\to C_{1}(X)$. 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 $u\in\ker(\pi)$ 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 $\overline{\pi}\colon H_{0}(X)=C_{0}(X)/B_{0}(X)\to{\mathbb{Z}}\{\pi_{0}(X)\}$ given by $\overline{\pi}(u+B_{0}(X))=\pi(u)$ for all $u\in C_{0}(X)$, and this is in fact an isomorphism. ∎

###### Example 10.26.

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 $\pi\circ\sigma=\operatorname{id}$. 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 $\ker(\pi\colon C_{0}(X)\to{\mathbb{Z}}\{\pi_{0}(X)\})$ could be the element $u=a_{1}-a_{2}+b_{1}+b_{3}-2b_{2}$. This has $\gamma(u)=\gamma(a_{1})-\gamma(a_{2})+\gamma(b_{1})+\gamma(b_{3})-2\gamma(b_{2})$, so

 $\displaystyle\partial(\gamma(u))$ $\displaystyle=(a_{1}-\sigma(A))-(a_{2}-\sigma(A))+(b_{1}-\sigma(B))+(b_{3}-% \sigma(B))-2(b_{2}-\sigma(B))$ $\displaystyle=a_{1}-a_{2}+b_{1}+b_{3}-2b_{2}=u,$

so $u=\partial(\gamma(u))\in\operatorname{img}(\partial)=B_{0}(X)$. This illustrates the fact that $\ker(\pi)=\operatorname{img}(\partial)$, which is a key step in our proof of Proposition 10.25.

We next discuss homology classes of paths, revisiting Remark 10.10.

###### Lemma 10.27.

Let $X$ be a topological space.

• (a)

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 $C_{1}(X)/B_{1}(X)$.

• (b)

For any path $u\colon a\rightsquigarrow b$ in $X$ with reversed path $\overline{u}\colon b\rightsquigarrow a$, we have $u+\overline{u}\in B_{1}(X)$ so $\overline{u}+B_{1}(X)=-u+B_{1}(X)$ in $C_{1}(X)/B_{1}(X)$.

• (c)

For any paths $u\colon a\rightsquigarrow b$ and $v\colon b\rightsquigarrow c$ we have $(u*v)+B_{1}(X)=(u+B_{1}(X))+(v+B_{1}(X))$ in $C_{1}(X)/B_{1}(X)$.

###### Proof.

Exercise.

Video (Path homotopy, loop homotopy and homology)

Video (Definition 10.28 and Proposition 10.29)

###### Definition 10.28.

Let $X$ be a topological space. A loop in $X$ is a path $u\colon\Delta_{1}\to X$ 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 $a$.

###### Proposition 10.29.

Let $X$ be a path connected space, and let $a$ be a point in $X$. Then for every $h\in H_{1}(X)$ there exists a loop $u$ based at $a$ with $h=[u]$. Moreover, if $u$ and $v$ are loops based at $a$ then so are $c_{a}$, $\overline{u}$ and $u*v$, and we have $[c_{a}]=0$ and $[\overline{u}]=-[u]$ and $[u*v]=[u]+[v]$ in $H_{1}(X)$.

###### Proof.

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

 $h=u_{1}+\dotsb+u_{n}+B_{1}(X)$

for some list of paths $u_{i}$. As this is a homology class, the representing chain must be a cycle, so we must have $\partial(u_{1}+\dotsb+u_{n})=0$ in $C_{0}(X)$. As $\partial(u_{i})=u_{i}(e_{1})-u_{i}(e_{0})$, this means that

 $u_{1}(e_{1})+\dotsb+u_{n}(e_{1})=u_{1}(e_{0})+\dotsb+u_{n}(e_{0}).$

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 $\{1,\dotsc,n\}$ 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 $(i\;j\;k\;l)$, 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. ∎

###### Definition 10.30.

Let $u\colon\Delta_{1}\to X$ be a loop based at $a$. A filling in of $u$ is a map $v\colon\Delta_{2}\to X$ with $v\circ\delta_{0}=u$ and $v\circ\delta_{1}=v\circ\delta_{2}=c_{a}$.

###### Lemma 10.31.

If $u$ can be filled in, then $[u]=0$ in $H_{1}(X)$.

###### Proof.

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 $[u]=u+B_{1}(X)=0$. ∎