MAS61015 Algebraic Topology 9 Homotopy 11 Homology of the punctured plane

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:

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

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

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

$\partial(u)=(a_{1}-a_{0})+(a_{2}-a_{1})+(a_{3}-a_{2})+(a_{4}-a_{3})+(a_{5}-a_{% 4})+(a_{6}-a_{5})=a_{6}-a_{0}.$
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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 $\partial\colon C_{2}(X)\to C_{1}(X)$ .
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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 }j<i\\ 0&\text{ if }j=i\\ t_{j-1}&\text{ if }j>i.\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.

Interactive demo

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.

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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.

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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 $0\leq i<j\leq k$ 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

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

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 $i<j$ . 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<j\leq k}u\circ\delta_{i}\circ\delta_{j-1}.

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

∎

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 $[u]=u+B_{1}(X)=0$ . ∎