Part II Lecture Notes: Algebraic Topology, James Lingard
Orientations
Let $\Delta$ be an $n$-simplex with vertices $x_{1}, \ldots, x_{n+1}$. If $n \geq 0$ then an orientation of $\Delta$ is an equivalence class of orderings of the $x_{i}$, where the orderings $x_{1}, \ldots, x_{n+1}$ and $x_{\sigma(1)}, \ldots, x_{\sigma(n+1)}$ are equivalent if and only if $\sigma \in A_{n+1}$. If $n=0$ then an orientation of $\Delta$ is either 1 or -1 .
An oriented simplex $\sigma$ is a simplex together with a choice of orientation. We write $-\sigma$ to mean the simplex with the opposite choice of orientation. We shall write $\left[x_{0}, \ldots, x_{n}\right]$ to mean the oriented simplex with the orientation corresponding to the given ordering of the vertices.
An oriented simplex induces an orientation on each of its codimension-one faces in the following manner:
- if $i$ is even, the induced orientation of $\left[x_{0}, \ldots, \hat{x}_{i}, \ldots, x_{n}\right]$ is the one corresponding to this ordering of the vertices, and
- if $i$ is odd, the induced orientation is the opposite to the one corresponding to this ordering.
We can check that this is well-defined.
Chains and boundaries
Let $X$ be a simplicial complex. We define the group $C_{k}(X)$ of $k$-chains on $X$ to be the free abelian group generated by the oriented $k$-simplices in $X$, modulo the relation that $(-1) \sigma=-\sigma$ for any oriented $k$-simplex $\sigma$.
For any oriented $k$-simplex $\sigma$ in $X$, we define the boundary $\partial \sigma$ of $\sigma$ to be the sum of the codimension-one faces of $\sigma$, with the orientations induced from $\sigma$. So $\partial \sigma \in C_{k-1}(X)$. We can extend this definition to get a group homomorphism $\partial: C_{k}(X) \rightarrow C_{k-1}$ in the obvious manner.
So we get the sequence
$$
\cdots \stackrel{\partial}{\longrightarrow} C_{k+1}(X) \stackrel{\partial}{\longrightarrow} C_{k}(X) \stackrel{\partial}{\longrightarrow} C_{k-1}(X) \stackrel{\partial}{\longrightarrow} \cdots
$$
In fact, the composition of any two of the consecutive homomorphisms is the zero homomorphism, or in other words, $\partial^{2}=0$. It suffices to prove that $\partial^{2} \sigma=0$ for any oriented $(k+1)$-simplex $\sigma$. We have
$$
\begin{aligned}
\partial^{2}\left[x_{0}, \ldots, x_{k+1}\right]= & \partial \sum_{i=0}^{k+1}(-1)^{i}\left[x_{0}, \ldots, \hat{x}_{i}, \ldots, x_{k+1}\right] \\
= & \sum_{i=0}^{k+1}(-1)^{i} \sum_{j=0}^{i-1}(-1)^{j}\left[x_{0}, \ldots, \hat{x}_{j}, \ldots, \hat{x}_{i}, \ldots, x_{k+1}\right] \\
& +\sum_{i=0}^{k+1}(-1)^{i} \sum_{j=i+1}^{k+1}(-1)^{j-1}\left[x_{0}, \ldots, \hat{x}_{i}, \ldots, \hat{x}_{j}, \ldots, x_{k+1}\right],
\end{aligned}
$$
where the terms in the final expression cancel in pairs.
Homology groups
Define the subgroup $Z_{k}(X) \leq C_{k}(X)$ of $k$-chains in $X$ to be the kernel of $\partial$, and define the subgroup $B_{k}(X) \leq C_{k}(X)$ of $k$-boundaries to be the image of $\partial$. Then since $\partial^{2}=0, B_{k}(X)$ is contained within $Z_{k}(X)$ and so we may define the $k$ th homology group of $X$ as the quotient
$$
H_{k}(X)=Z_{k}(X) / B_{k}(X)
$$
It is easy to see that $H_{0}(X)$ is a free abelian group whose rank is the number of connected components of $X$.
Homology groups of an $n$-simplex
If $X$ is a point, then by considering the groups of $k$-chains on $X$ we see immediately that $H_{0}(X)=\mathbb{Z}$ and $H_{i}(X)=0$ for $i \geq 1$.
To compute the homology groups of an $n$-simplex for $n \geq 2$ we may use the following device. If $X \subset \mathbb{R}^{N}$ is a simplicial complex, define the cone on $X$ to be the simplicial complex $C X$ formed by embedding $\mathbb{R}^{N}$ in $\mathbb{R}^{N+1}$, choosing a point $v \in \mathbb{R}^{N+1} \backslash \mathbb{R}^{N}$ and then taking the union of all the line segments from $v$ to each of the points in $X$. Then we see that $C X$ is a simplicial complex, whose simplices are those in $X$, those formed as the convex hull of $v$ and a simplex in $X$, and the point $v$ itself.
The homology groups of any cone are the same as those of a point. To prove this we define the homomorphism $d: C_{q}(C X) \rightarrow C_{q+1}(C X)$ which sends an oriented $q$-simplex $\sigma=\left[v_{0}, \ldots, v_{q}\right]$ in $C X$ to $\left[v, v_{0}, \ldots, v_{q}\right]$ if $\sigma$ is contained in $X$, or 0 otherwise. We can check that this is a well-defined homomorphism. Then we show that
$$
(\partial \circ d)(\sigma)=\sigma-(d \circ \partial)(\sigma)
$$
for any oriented $q$-simplex $\sigma$ with $q \geq 1$. But then if $z$ is any $q$-cycle in $C X$ with $q \geq 1$,
$$
\partial(d(z))=z-d(\partial(z))=z-d(0)=z
$$
and so $z$ is a boundary. Hence $H_{q}(C X)=0$.
Now since an $(n+1)$-simplex is just the cone on an $n$-simplex, we have that for any simplex $X$, $H_{0}(X)=\mathbb{Z}$ and $H_{i}(X)=0$ for $i \geq 1$.
Homology groups of $S^{n}$
If $n=0$ then the homology groups of $S^{n}$ are clearly $H_{0}\left(S^{0}\right)=Z^{2}, H_{i}\left(S^{0}\right)=0$ for $i \geq 1$.
If $n \geq 1$ then we may triangulate $S^{n}$ is a simple way. Let $\Delta$ denote the simplicial complex which is the union of all the faces of an $(n+1)$-simplex, and let $\Sigma$ denote the same simplicial complex but without the single $(n+1)$-dimensional face. Then $\Sigma$ is a triangulation of $S^{n}$. Now the chain complex of $\Sigma$ is exactly the same as that of $\Delta$, except that we have $C_{n+1}(\Sigma)=0$, whereas $C_{n+1}(\Delta) \cong \mathbb{Z}$, generated by the single $(n+1)$-dimensional face.
Therefore, for $i \leq(n-1)$ the homology groups $H_{i}(\Delta)$ and $H_{i}(\Sigma)$ are the same. But then we see that $H_{n}(\Sigma)$ is the kernel of the map $\partial: C_{n}(\Sigma) \rightarrow C_{n-1}(\Sigma)$, which is also the kernel of $\partial: C_{n}(\Delta) \rightarrow C_{n-1}(\Delta)$. But by the exactness of the sequence for $\Delta$, this is just the image of $\partial: C_{n+1}(\Delta) \rightarrow C_{n}(\Delta)$, which is isomorphic to $C_{n+1}(\Delta) \cong \mathbb{Z}$ as $\partial$ is injective.
We will show that the homology groups of a space are independent of its triangulation, and hence we may conclude that
$$
H_{i}\left(S^{n}\right)= \begin{cases}\mathbb{Z}^{2} & \text { if } n=0 \text { and } i=0 \\ \mathbb{Z} & \text { if } n \geq 1 \text { and } i=0 \text { or } n \\ 0 & \text { otherwise. }\end{cases}
$$
Induced homomorphisms
Let $f: X \rightarrow Y$ be a simplicial map. Then $f$ determines a homomorphism
$$
f_{*}: C_{i}(X) \longrightarrow C_{i}(Y)
$$
by
$$
f_*([x_{0}, \ldots, x_{i}])= \begin{cases}{[f(x_0), \dots, f(x_i)]} & \text { if } f(x_0), \ldots, f(x_i) \text { are all distinct } \\ 0 & \text { otherwise }\end{cases}
$$
We can check that $f_{*}$ is a chain map, that is, that
$$
\partial \circ f_{*}=f_{*} \circ \partial
$$
and so the following diagram commutes:
\begin{CD}
\cdots@>\partial>>C_{i+1}(X)@>\partial>>C_i(X)@>\partial>>C_{i-1}(X)@>\partial>>\cdots\\
&@VVf_*V@VVf_*V@VVf_*V\\
\cdots@>\partial>>C_{i+1}(Y)@>\partial>>C_i(Y)@>\partial>>C_{i-1}(Y)@>\partial>>\cdots
\end{CD}
It follows that $f_{*}$ maps $Z_{i}(X)$ into $Z_{i}(Y)$, and $B_{i}(X)$ into $B_{i}(Y)$, and so $f_{*}$ gives rise to a homomorphism, also denoted $f_{*}$, from $H_{i}(X)$ to $H_{i}(Y)$.
Now it can be shown that barycentric subdivision of a complex does not change the homology groups (proof omitted - see Armstrong pp. 185-188). Thus if $f:|X| \rightarrow|Y|$ is any continuous map then we may define a homomorphism
$$
f_{*}: H_{i}(X) \rightarrow H_{i}(Y)
$$
as the composition
$$
H_{i}(X) \cong H_{i}\left(X^{m}\right) \stackrel{s_{*}}{\longrightarrow} H_{i}(Y)
$$
where $s: X \rightarrow Y$ is some simplicial approximation to $f$. It can be shown that this is welldefined, independent of the choice of $s$, by showing that any two "close" simplicial maps give rise to the same homomorphisms of homology groups.
Functorial properties. Thus homology groups are invariant under homotopy equivalence.
Applications
$S^{m}$ and $S^{n}$ are not homotopy equivalent and thus $\mathbb{R}^{m}$ and $\mathbb{R}^{n}$ are not homeomorphic if $m \neq n$. The Brouwer fixed-point theorem.