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algebraic_topology_iii.tex
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algebraic_topology_iii.tex
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\documentclass[a4paper]{article}
\def\npart {III}
\def\nterm {Michaelmas}
\def\nyear {2016}
\def\nlecturer {O.\ Randal-Williams}
\def\ncourse {Algebraic Topology}
\input{header}
\theoremstyle{definition}
\newtheorem{cclaim}{Claim}
\setcounter{cclaim}{-1}
\begin{document}
\maketitle
{\small
\setlength{\parindent}{0em}
\setlength{\parskip}{1em}
Algebraic Topology assigns algebraic invariants to topological spaces; it permeates modern pure mathematics. This course will focus on (co)homology, with an emphasis on applications to the topology of manifolds. We will cover singular homology and cohomology, vector bundles and the Thom Isomorphism theorem, and the cohomology of manifolds up to Poincar\'e duality. Time permitting, there will also be some discussion of characteristic classes and cobordism, and conceivably some homotopy theory.
\subsubsection*{Pre-requisites}
Basic topology: topological spaces, compactness and connectedness, at the level of Sutherland's book. The course will not assume any knowledge of Algebraic Topology, but will go quite fast in order to reach more interesting material, so some previous exposure to simplicial homology or the fundamental group would be helpful. The Part III Differential Geometry course will also contain useful, relevant material.
Hatcher's book is especially recommended for the course, but there are many other suitable texts.
}
\tableofcontents
\section{Homotopy}
In this course, the word ``map'' will always mean ``continuous function''.
In topology, we study spaces up to ``continuous deformation''. Famously, a coffee mug can be continuously deformed into a doughnut, and thus they are considered to be topologically the same. Now we also talk about maps between topological spaces. So a natural question is if it makes sense to talk about the continuous deformations of maps. It turns out it does, and the definition is sort-of the obvious one:
\begin{defi}[Homotopy]\index{homotopy}
Let $X, Y$ be topological spaces. A \emph{homotopy} between $f_0, f_1: X \to Y$ is a map $F: [0, 1] \times X \to Y$ such that $F(0, x) = f_0(x)$ and $F(1, x) = f_1(x)$. If such an $F$ exists, we say $f_0$ is \emph{homotopic} to $f_1$, and write $f_0 \simeq f_1$.
This $\simeq$ defines an equivalence relation on the set of maps from $X$ to $Y$.
\end{defi}
Just as concepts in topology are invariant under homeomorphisms, it turns out the theories we develop in algebraic topology are invariant under homotopies, i.e.\ homotopic maps ``do the same thing''. This will be made more precise later.
Under this premise, if we view homotopic functions as ``the same'', then we have to enlarge our notion of isomorphism to take this into account. To do so, we just write down the usual definition of isomorphism, but with equality replaced with homotopy.
\begin{defi}[Homotopy equivalence]\index{homotopy equivalence}
A map $f: X \to Y$ is a \emph{homotopy equivalence} if there is some $g: Y \to X$ such that $f \circ g \simeq \id_Y$ and $g \circ f \simeq \id_X$. We call $g$ a \emph{homotopy inverse}\index{homotopy inverse} to $f$.
\end{defi}
As always, homotopy equivalence is an equivalence relation. This relies on the following straightforward property of homotopy:
\begin{prop}
If $f_0 \simeq f_1: X \to Y$ and $g_0 \simeq g_1: Y \to Z$, then $g_0 \circ f_0 \simeq g_1 \circ f_1: X \to Z$.
\[
\begin{tikzcd}
X \ar[r, bend left, "f_0"] \ar[r, bend right, "f_1"'] & Y \ar[r, bend left, "g_0"] \ar[r, bend right, "g_1"'] & Z
\end{tikzcd}
\]
\end{prop}
\begin{eg}[Stupid example]
If $f: X \to Y$ is a homeomorphism, then it is a homotopy equivalence --- we take the actual inverse for the homotopy inverse, since equal functions are homotopic.
\end{eg}
\begin{eg}[Interesting example]
Let $i: \{0\} \to \R^n$ be the inclusion map. To show this is a homotopy equivalence, we have to find a homotopy inverse. Fortunately, there is only one map $\R^n \to \{0\}$, namely the constant function $0$. We call this $r: \R^n \to \{0\}$. The composition $r \circ i: \{0\} \to \{0\}$ is exactly the identity. So this is good.
In the other direction, the map $i \circ r: \R^n \to \R^n$ sends everything to $0$. We need to produce a homotopy to the identity. We let $F: [0, 1] \times \R^n \to \R^n$ be
\[
F(t, \mathbf{v}) = t\mathbf{v}.
\]
We have $F(0, \mathbf{v}) = 0$ and $F(1, \mathbf{v}) = \mathbf{v}$. So this is indeed a homotopy from $i \circ r$ to $\id_{\R^n}$.
\end{eg}
So from the point of view of homotopy, the one-point space $\{0\}$ is the same as $\R^n$! So dimension, or even cardinality, is now a meaningless concept.
\begin{eg}[Also interesting example]
Let $S^n \subseteq \R^{n + 1}$ be the unit sphere, and $i: S^n \hookrightarrow \R^{n + 1} \setminus \{0\}$. We show that this is a homotopy equivalence. We define $r: \R^{n + 1} \setminus \{0\} \to S^n$ by
\[
r(\mathbf{v}) = \frac{\mathbf{v}}{\|\mathbf{v}\|}.
\]
Again, we have $r \circ i = \id_{S^n}$. In the other direction, we need to construct a path from each $\mathbf{v}$ to $\frac{\mathbf{v}}{\|\mathbf{v}\|}$ in a continuous way. We could do so by
\begin{align*}
H: [0, 1] \times (\R^{n + 1} \setminus \{0\}) &\to \R^{n + 1} \setminus \{0\}\\
(t, \mathbf{v}) &\mapsto (1 - t) \mathbf{v} + t \frac{\mathbf{v}}{\|\mathbf{v}\|}.
\end{align*}
We can easily check that this is a homotopy from $\id_{\R^{n + 1}\setminus \{0\}}$ to $i \circ r$.
\end{eg}
Again, homotopy equivalence allowed us to squash down one dimension of $\R^{n + 1} \setminus \{0\}$ to get $S^n$. However, there is one thing we haven't gotten rid of --- the hole. It turns out what \emph{is} preserved by homotopy equivalence is exactly the holes.
Now we might ask ourselves --- can we distinguish holes of ``different dimension''? For $n \not= m$, is $S^n$ homotopy equivalent to $S^m$? If we try hard to construct homotopies, we will always fail, and then we would start to think that maybe they aren't homotopy equivalent. However, at this point, we do not have any tools we can use the prove this.
The solution is algebraic topology. The idea is that we assign algebraic objects to each topological space in a way that is homotopy-invariant. We then come up with tools to compute these algebraic invariants. Then if two spaces are assigned different algebraic objects, then we know they cannot be homotopy equivalent.
What is this good for? At this point, you might not be convinced that it is a good idea to consider spaces up to homotopy equivalence only. However, algebraic topology can also help us show that spaces are not homeomorphic. After all, homeomorphic spaces are also homotopy equivalent. For example, suppose we want to show that if $\R^n \cong \R^m$, then $n = m$. Algebraic topology doesn't help us directly, because both $\R^n$ and $\R^m$ are homotopy equivalent to a point. Instead, we do the slightly sneaky thing of removing a point. If $\R^n \cong \R^m$, then it must also be the case that $\R^n \setminus \{0\} \cong \R^m \setminus \{0\}$. Since these are homotopy equivalent to $S^{n - 1}$ and $S^{m - 1}$, this implies that $S^{n - 1} \simeq S^{m - 1}$ are homotopy equivalent. By algebraic topology, we will show that that this can only happen if $m = n$. So indeed we can recover the notion of dimension using algebraic topology!
\section{Singular (co)homology}
\subsection{Chain complexes}
This course is called algebraic topology. We've already talked about some topology, so let's do some algebra. We will just write down a bunch of definitions, which we will get to use in the next chapter to define something useful.
\begin{defi}[Chain complex]\index{chain complex}
A \emph{chain complex} is a sequence of abelian groups and homomorphisms
\[
\begin{tikzcd}
\cdots \ar[r] & C_3 \ar[r, "d_3"] & C_2 \ar[r, "d_2"] & C_1 \ar[r, "d_1"] & C_0 \ar[r, "d_0"] & 0
\end{tikzcd}
\]
such that
\[
d_i \circ d_{i + 1} = 0
\]
for all $i$.
\end{defi}
Very related to the notion of a chain complex is a \emph{co}chain complex, which is the same thing with the maps the other way.
\begin{defi}[Cochain complex]\index{cochain complex}
A \emph{cochain complex} is a sequence of abelian groups and homomorphisms
\[
\begin{tikzcd}
0 \ar[r] & C^0 \ar[r, "d^0"] & C^1 \ar[r, "d^1"] & C^2 \ar[r, "d^2"] & C^3 \ar[r] & \cdots
\end{tikzcd}
\]
such that
\[
d^{i + 1} \circ d^i = 0
\]
for all $i$.
\end{defi}
Note that each of the maps $d$ is indexed by the number of its domain.
\begin{defi}[Differentials]\index{differentials}
The maps $d^i$ and $d_i$ are known as \emph{differentials}.
\end{defi}
Eventually, we will get lazy and just write all the differentials as $d$.
Given a chain complex, the only thing we know is that the composition of any two maps is zero. In other words, we have $\im d_{i+1} \subseteq \ker d_i$. We can then ask how good this containment is. Is it that $\im d_{i + 1} = \ker d_i$, or perhaps that $\ker d_i$ is huge but $\im d_{i + 1}$ is trivial? The homology or cohomology measures what happens.
\begin{defi}[Homology]\index{homology}
The \emph{homology} of a chain complex $C_{\Cdot}$ is
\[
H_i(C_{\Cdot}) = \frac{\ker (d_i: C_i \to C_{i - 1})}{\im (d_{i + 1}: C_{i + 1} \to C_i)}.
\]
An element of $H_i(C_{\Cdot})$ is known as a \term{homology class}.
\end{defi}
Dually, we have
\begin{defi}[Cohomology]\index{cohomology}
The \emph{cohomology} of a cochain complex $C^{\Cdot}$ is
\[
H^i(C^{\Cdot}) = \frac{\ker (d^i: C^i \to C^{i + 1})}{\im (d^{i - 1}: C^{i - 1} \to C^i)}.
\]
An element of $H^i(C^{\Cdot})$ is known as a \term{cohomology class}.
\end{defi}
More names:
\begin{defi}[Cycles and cocycles]\index{cycle}\index{cocycle}
The elements of $\ker d_i$ are the \emph{cycles}, and the elements of $\ker d^i$ are the \emph{cocycles}.
\end{defi}
\begin{defi}[Boundaries and coboundaries]\index{boundary}\index{coboundary}
The elements of $\im d_i$ are the \emph{boundaries}, and the elements of $\im d^i$ are the \emph{coboundaries}.
\end{defi}
As always, we want to talk about maps between chain complexes. These are known as chain maps.
\begin{defi}[Chain map]\index{chain map}
If $(C_{\Cdot}, d_{\Cdot}^C)$ and $(D_{\Cdot}, d_{\Cdot}^D)$ are chain complexes, then a \emph{chain map} $C_{\Cdot} \to D_{\Cdot}$ is a collection of homomorphisms $f_n: C_n \to D_n$ such that $d_n^D \circ f_n = f_{n - 1} \circ d_n^C$. In other words, the following diagram has to commute for all $n$:
\[
\begin{tikzcd}
C_n \ar[r, "f_n"] \ar[d, "d_n^C"] & D_n \ar[d, "d_n^D"]\\
C_{n - 1} \ar[r, "f_{n - 1}"] & D_{n - 1}
\end{tikzcd}
\]
\end{defi}
There is an obvious analogous definition for \term{cochain maps} between cochain complexes.
\begin{lemma}
If $f_{\Cdot}: C_{\Cdot} \to D_{\Cdot}$ is a chain map, then $f_*: H_n(C_{\Cdot}) \to H_n(D_{\Cdot})$ given by $[x] \mapsto [f_n(x)]$ is a well-defined homomorphism, where $x \in C_n$ is any element representing the homology class $[x] \in H_n(C_{\Cdot})$.
\end{lemma}
\begin{proof}
Before we check if it is well-defined, we first need to check if it is defined at all! In other words, we need to check if $f_n(x)$ is a cycle. Suppose $x \in C_n$ is a cycle, i.e.\ $d_n^C(x) = 0$. Then we have
\[
d_n^D(f_n(x)) = f_{n - 1}(d_n^C(x)) = f_{n - 1}(0) = 0.
\]
So $f_n(x)$ is a cycle, and it does represent a homology class.
To check this is well-defined, if $[x] = [y] \in H_n(C_{\Cdot})$, then $x - y = d_{n + 1}^C(z)$ for some $z\in C_{n + 1}$. So $f_n(x) - f_n(y) = f_n (d_{n + 1}^C(z)) = d_{n + 1}^D (f_{n + 1}(z))$ is a boundary. So we have $[f_n(x)] = [f_n(y)] \in H_n(D_{\Cdot})$.
\end{proof}
\subsection{Singular (co)homology}
The idea now is that given any space $X$, we construct a chain complex $C_\Cdot(X)$, and for any map $f: X \to Y$, we construct a chain map $f_\Cdot: C_\Cdot(X) \to C_\Cdot(Y)$. We then take the homology of these, so that we get some homology groups $H_*(X)$ for each space $X$, and a map $f_*: H_*(X) \to H_*(Y)$ for each map $f: X \to Y$.
There are many ways we can construct a chain complex $C_\Cdot(X)$ from a space $X$. The way we are going to define the chain complex is via \emph{singular homology}. The advantage of this definition is that it is obviously a property just of the space $X$ itself, whereas in other definitions, we need to pick, say, a triangulation of the space, and then work hard to show that the homology does not depend on the choice of the triangulation.
The disadvantage of singular homology is that the chain complexes $C_\Cdot(X)$ will be \emph{huge} (and also a bit scary). Except for the case of a point (or perhaps a few points), it is impossible to actually write down what $C_\Cdot(X)$ looks like and use that to compute the homology groups. Instead, we are going to play around with the definition and prove some useful results that help us compute the homology groups indirectly.
Later, for a particular type of spaces known as \emph{CW complexes}, we will come up with a different homology theory where the $C_{\Cdot}(X)$ are actually nice and small, so that we can use it to compute the homology groups directly. We will prove that this is equivalent to singular homology, so that this also provides an alternative method of computing homology groups.
Everything we do can be dualized to talk about cohomology instead. Most of the time, we will just write down the result for singular homology, but analogous results hold for singular cohomology as well. However, later on, we will see that there are operations we can perform on cohomology groups only, which makes them better, and the interaction between homology and cohomology will become interesting when we talk about manifolds at the end of the course.
We start with definitions.
\begin{defi}[Standard $n$-simplex]\index{standard $n$-simplex}
The standard $n$-simplex is
\[
\Delta^n = \left\{(t_0, \cdots, t_n) \in \R^{n + 1} : t_i \geq 0, \sum t_i = 1\right\}.
\]
\end{defi}
\begin{center}
\begin{tikzpicture}
\draw [->] (0, 0) -- (3, 0);
\draw [->] (0, 0) -- (0, 3);
\draw [->] (0, 0) -- (-1.5, -1.5);
\draw [fill=mblue, fill opacity=0.5] (0, 2) -- (2, 0) -- (-.866, -0.866) -- cycle;
\end{tikzpicture}
\end{center}
We notice that $\Delta^n$ has $n + 1$ ``faces''.
\begin{defi}[Face of standard simplex]\index{face of standard simplex}
The \emph{$i$th face} of $\Delta^n$ is
\[
\Delta_i^n = \{(t_0, \cdots, t_n) \in \Delta^n : t_i = 0\}.
\]
\end{defi}
\begin{eg}
The faces of $\Delta^1$ are labelled as follows:
\begin{center}
\begin{tikzpicture}
\draw [->] (-1, 0) -- (3, 0) node [right] {$t_0$};
\draw [->] (0, -1) -- (0, 3) node [above] {$t_1$};
\draw (2, 0) -- (0, 2) node [pos=0.5, anchor = south west] {$\Delta_1$};
\node [circ] at (2, 0) {};
\node [circ] at (0, 2) {};
\node at (2, 0) [below] {$\Delta_1^1$};
\node at (0, 2) [left] {$\Delta_0^1$};
\end{tikzpicture}
\end{center}
\end{eg}
We see that the $i$th face of $\Delta^n$ looks like the standard $(n-1)$-simplex. Of course, it is just homeomorphic to it, with the map given by
\begin{align*}
\delta_i: \Delta^{n - 1} &\to \Delta^n\\
(t_0, \cdots, t_{n - 1}) &\mapsto (t_0, \cdots, t_{i - 1}, 0, t_i, \cdots, t_{n - 1})
\end{align*}
This is a homeomorphism onto $\Delta_i^n$. We will make use of these maps to define our chain complex.
The idea is that we will look at subspaces of $X$ that ``look like'' these standard $n$-simplices.
\begin{defi}[Singular $n$-simplex]\index{singular $n$-simplex}
Let $X$ be a space. Then a \emph{singular $n$-simplex} in $X$ is a map $\sigma: \Delta^n \to X$.
\end{defi}
\begin{eg}
The inclusion of the standard $n$-simplex into $\R^{n+1}$ is a singular simplex, but so is the constant map to any point in $X$. So the singular $n$-simplices can be stupid.
\end{eg}
Given a space $X$, we obtain a \emph{set} of singular $n$-simplices. To talk about homology, we need to have an abelian group. We do the least imaginative thing ever:
\begin{defi}[Singular chain complex]
We let $C_n(X)$ be the free abelian group on the set of singular $n$-simplices in $X$. More explicitly, we have
\[
C_n(X) = \left\{\sum n_\sigma \sigma: \sigma: \Delta^n \to X, n_\sigma \in \Z, \text{only finitely many $n_\sigma$ non-zero}\right\}.
\]
We define $d_n: C_n(X) \to C_{n - 1}(X)$ by
\[
\sigma \mapsto \sum_{i = 0}^n (-1)^i \sigma \circ \delta_i,
\]
and then extending linearly.
\end{defi}
Note that it is essential that we insert those funny negative signs. Indeed, if the signs weren't there, then all terms in $d(d(\sigma))$ would have positive coefficients, and there is no hope that they all cancel. Intuitively, we can think of these signs as specifying the ``orientation'' of the faces. For example, if we have a line
\begin{center}
\begin{tikzpicture}
\draw (0, 0) node [circ] {} -- (2, 0) node [circ] {};
\end{tikzpicture}
\end{center}
then after taking $d$, one vertex would have a positive sign, and the other would have a negative sign.
Now we actually check that this indeed gives us a chain complex. The key tool is the following unexciting result:
\begin{lemma}
If $i < j$, then $\delta_j \circ \delta_i = \delta_i \circ \delta_{j - 1} : \Delta^{n - 2} \to \Delta^n$.
\end{lemma}
\begin{proof}
Both send $(t_0, \cdots, t_{n - 2})$ to
\[
(t_0, \cdots, t_{i - 1}, 0, t_i, \cdots, t_{j - 2}, 0, t_{j - 1}, \cdots, t_{n - 2}).\qedhere
\]
\end{proof}
\begin{cor}
The homomorphism $d_{n - 1} \circ d_n: C_n(X) \to C_{n - 2}(X)$ vanishes.
\end{cor}
\begin{proof}
It suffices to check this on each basis element $\sigma: \Delta^n \to X$. We have
\begin{align*}
d_{n - 1} \circ d_n (\sigma) &= \sum_{i = 0}^{n - 1}(-1)^i \sum_{j = 0}^n (-1)^j \sigma \circ \delta_j \circ \delta_i.
\intertext{We use the previous lemma to split the sum up into $i < j$ and $i \geq j$:}
&= \sum_{i < j} (-1)^{i + j} \sigma \circ \delta_j \circ \delta_i + \sum_{i \geq j} (-1)^{i + j} \sigma \circ \delta_j \circ \delta_i\\
&= \sum_{i < j} (-1)^{i + j} \sigma \circ \delta_i \circ \delta_{j - 1} + \sum_{i \geq j} (-1)^{i + j} \sigma \circ \delta_j \circ \delta_i\\
&= \sum_{i \leq j} (-1)^{i + j + 1} \sigma \circ \delta_i \circ \delta_j + \sum_{i \geq j} (-1)^{i + j} \sigma \circ \delta_j \circ \delta_i\\
&= 0.\qedhere
\end{align*}
\end{proof}
So the data $d_n: C_n (X) \to C_{n - 1}(X)$ is indeed a chain complex. The only thing we can do to a chain complex is to take its homology!
\begin{defi}[Singular homology]\index{singular homology}
The \emph{singular homology} of a space $X$ is the homology of the chain complex $C_\Cdot(X)$:
\[
H_i(X) = H_i(C_{\Cdot}(X), d_{\Cdot}) = \frac{\ker (d_i: C_i(X) \to C_{i - 1}(X))}{ \im(d_{i + 1}: C_{i + 1}(X) \to C_i(X))}.
\]
\end{defi}
We will also talk about the ``dual'' version of this:
\begin{defi}[Singular cohomology]\index{singular cohomology}
We define the dual cochain complex by
\[
C^n(X) = \Hom(C_n(X), \Z).
\]
We let
\[
d^n: C^n(X) \to C^{n + 1}(X)
\]
be the adjoint to $d_{n + 1}$, i.e.
\[
(\varphi: C_n(X) \to \Z) \mapsto (\varphi \circ d_{n + 1}: C_{n + 1}(X) \to \Z).
\]
We observe that
\[
\begin{tikzcd}
0 \ar[r] & C^0(X) \ar[r, "d^0"] & C^1(X) \ar[r] & \cdots
\end{tikzcd}
\]
is indeed a cochain complex, since
\[
d^{n + 1}(d^n(\varphi)) = \varphi \circ d_{n + 1} \circ d_{n + 2} = \varphi \circ 0 = 0.
\]
The \emph{singular cohomology} of $X$ is the cohomology of this cochain complex, i.e.
\[
H^i(X) = H^i(C^{\Cdot}(X), d^{\Cdot}) = \frac{\ker(d^i: C^i(X) \to C^{i + 1}(X))}{\im(d^{i - 1}: C^{i - 1}(X) \to C^i (X))}.
\]
\end{defi}
Note that in general, it is \emph{not} true that $H^n(X) = \Hom(H_n(X), \Z)$. Thus, dualizing and taking homology do not commute with each other. However, we will later come up with a relation between the two objects.
The next thing to show is that maps of spaces induce chain maps, hence maps between homology groups.
\begin{prop}
If $f: X \to Y$ is a continuous map of topological spaces, then the maps
\begin{align*}
f_n: C_n(X) &\to C_n(Y)\\
(\sigma: \Delta^n \to X) &\mapsto (f \circ \sigma: \Delta^n \to Y)
\end{align*}
give a chain map. This induces a map on the homology (and cohomology).
\end{prop}
\begin{proof}
To see that the $f_n$ and $d_n$ commute, we just notice that $f_n$ acts by composing on the left, and $d_n$ acts by composing on the right, and these two operations commute by the associativity of functional composition.
\end{proof}
Now if in addition we have a map $g: Y \to Z$, then we obtain two maps $H_n(X) \to H_n(Z)$ by
\[
\begin{tikzcd}
H_n(X) \ar[rr, "(g \circ f)_*"] \ar[rd, "f_*"] & & H_n(Z)\\
& H_n(Y) \ar[ru, "g_*"]
\end{tikzcd}.
\]
By direct inspection of the formula, we see that this diagram commutes. In equation form, we have
\[
(g \circ f)_* = g_* \circ f_*.
\]
Moreover, we trivially have
\[
(\id_X)_* = \id_{H_n(X)}: H_n(X) \to H_n(X).
\]
Thus we deduce that
\begin{prop}
If $f: X \to Y$ is a homeomorphism, then $f_*: H_n(X) \to H_n(Y)$ is an isomorphism of abelian groups.
\end{prop}
\begin{proof}
If $g: Y \to X$ is an inverse to $f$, then $g_*$ is an inverse to $f_*$, as $f_* \circ g_* = (f \circ g)_* = (\id)_* = \id$, and similarly the other way round.
\end{proof}
If one is taking category theory, what we have shown is that $H_*$ is a functor, and the above proposition is just the usual proof that functors preserve isomorphisms.
This is not too exciting. We will later show that \emph{homotopy equivalences} induce isomorphisms of homology groups, which is much harder.
Again, we can dualize this to talk about cohomology. Applying $\Hom(\ph, \Z)$ to $f_{\Cdot}: C_{\Cdot}(X) \to C_{\Cdot}(Y)$ gives homomorphisms $f^n: C^n(Y) \to C^n(X)$ by mapping
\[
(\varphi: C_n(Y) \to \Z) \mapsto (\varphi \circ f_n: C_n(X) \to \Z).
\]
Note that this map goes the other way! Again, this is a cochain map, and induces maps $f^*: H^n(Y) \to H^n(X)$.
How should we think about singular homology? There are two objects involved --- cycles and boundaries. We will see that cycles in some sense ``detect holes'', and quotienting out by boundaries will help us identify cycles that detect the same hole.
\begin{eg}
We work with the space that looks like this:
\begin{center}
\begin{tikzpicture}
\draw [fill=morange, opacity=0.3] circle [radius=1.5];
\draw circle [radius=1.5];
\draw [fill=white] circle [radius=0.5];
\end{tikzpicture}
\end{center}
Suppose we have a single chain complex $\sigma: \Delta^1 \to X$. Then its boundary $d_1(\sigma) = \sigma(1) - \sigma(0)$. This is in general non-zero, unless we have $\sigma(0) = \sigma(1)$. In other words, this has to be a loop!
\begin{center}
\begin{tikzpicture}
\draw [fill=morange, opacity=0.3] circle [radius=1.5];
\draw circle [radius=1.5];
\draw [fill=white] circle [radius=0.5];
\draw [mblue, thick] (-1, 0) to [out=-80, in=270, looseness=1.2] (0.8, 0) node [right] {$\sigma$} to [out=90, in=80, looseness=1.2] (-1, 0) node [circ] {};
\end{tikzpicture}
\end{center}
So the $1$-cycles represented by just one element are exactly the loops.
How about more complicated cycles? We could have, say, four $1$-simplices $\sigma_1, \sigma_2, \sigma_3, \sigma_4$.
\begin{center}
\begin{tikzpicture}
\draw [fill=morange, opacity=0.3] circle [radius=1.5];
\draw circle [radius=1.5];
\draw [fill=white] circle [radius=0.5];
\draw [mblue, thick] (-1, 0) to [out=-80, in=180] (0, -0.65) node [circ] {};
\draw [mgreen, thick] (0, -0.65) to [out=0, in=270] (0.8, 0) node [circ] {};
\draw [mred, thick] (0.8, 0) to [out=90, in=0] (0, 0.65) node [circ] {};
\draw [morange, thick] (0, 0.65) to [out=180, in=80] (-1, 0) node [circ] {};
\node [circ] at (-1, 0) {};
\node [circ] at (0, -0.65) {};
\node [circ] at (0.8, 0) {};
\node [circ] at (0, 0.65) {};
\node [mblue] at (-0.8, -0.7) {$\sigma_1$};
\node [mgreen] at (0.8, -0.7) {$\sigma_2$};
\node [mred] at (0.8, 0.7) {$\sigma_3$};
\node [morange] at (-0.8, 0.7) {$\sigma_4$};
\end{tikzpicture}
\end{center}
In this case we have
\begin{align*}
\sigma_1(1) &= \sigma_2(0)\\
\sigma_2(1) &= \sigma_3(0)\\
\sigma_3(1) &= \sigma_4(0)\\
\sigma_4(1) &= \sigma_1(0)
\end{align*}
Thus, we have $\sigma_1 + \sigma_2 + \sigma_3 + \sigma_4 \in C_1(X)$. We can think of these cycles as detecting the holes by surrounding them.
However, there are some stupid cycles:
\begin{center}
\begin{tikzpicture}
\draw [fill=morange, opacity=0.3] circle [radius=1.5];
\draw circle [radius=1.5];
\draw [fill=white] circle [radius=0.5];
\draw [mblue, thick] (0.7, 0) to [out=-80, in=270, looseness=1.2] (1.3, 0) to [out=90, in=80, looseness=1.2] (0.7, 0) node [circ] {};
\end{tikzpicture}
\end{center}
These cycles don't really surround anything. The solution is that these cycles are actually boundaries, as it is the boundary of the $2$-simplex that fills the region bounded by the loop (details omitted).
Similarly, the boundaries also allow us to identify two cycles that surround the same hole, so that we don't double count:
\begin{center}
\begin{tikzpicture}
\draw [fill=morange, opacity=0.3] circle [radius=1.5];
\draw circle [radius=1.5];
\draw [fill=white] circle [radius=0.5];
\draw [mblue, thick] (-0.8, 0) to [out=-80, in=270, looseness=1.4] (0.6, 0) to [out=90, in=80, looseness=1.4] (-0.8, 0) node [circ] {};
\draw [mgreen, thick] (-1.3, 0) to [out=-80, in=270, looseness=1.2] (1.1, 0) to [out=90, in=80, looseness=1.2] (-1.3, 0) node [circ] {};
\end{tikzpicture}
\end{center}
This time, the \emph{difference} between the two cycles is the boundary of the $2$-simplex given by the region in between the two loops.
Of course, some work has to be done to actually find a $2$-simplex whose boundary is the difference of the two loops above, and in fact we will have to write the region as the sum of multiple $2$-simplices for this to work. However, this is just to provide intuition, not a formal proof.
\end{eg}
We now do some actual computations. If we want to compute the homology groups directly, we need to know what the $C_\Cdot(X)$ look like. In general, this is intractable, unless we have the very simple case of a point:
\begin{eg}
Consider the one-point space $\pt = \{*\}$. We claim that
\[
H^n(\pt) = H_n(\pt) =
\begin{cases}
\Z & n = 0\\
0 & n > 0
\end{cases}.
\]
To see this, we note that there is always a single singular $n$-simplex $\sigma_n: \Delta^n \to \pt$. So $C_n(\pt) = \Z$, and is generated by $\sigma_n$. Now note that
\[
d_n(\sigma_n) = \sum_{i = 0}^n (-1)^i \sigma_n \delta_i =
\begin{cases}
\sigma_{n - 1} & n\text{ even}\\
0 & n\text{ odd}
\end{cases}.
\]
So the singular chain complex looks like
\[
\begin{tikzcd}
\Z \ar[r, "1"] & \Z\ar[r, "0"] & \Z \ar[r, "1"] & \Z \ar[r, "0"] & \Z \ar[r] & 0.
\end{tikzcd}
\]
The homology groups then follow from direct computation. The cohomology groups are similar.
\end{eg}
This result is absolutely unexciting, and it is also almost the only space whose homology groups we can find at this point. However, we are still capable of proving the following general result:
\begin{eg}
If $X = \coprod_{\alpha \in I} X_\alpha$ is a disjoint union of path components, then each singular simplex must lie in some $X_\alpha$. This implies that
\[
H_n(X) \cong \bigoplus_{\alpha \in I} H_n(X_\alpha).
\]
\end{eg}
Now we know how to compute, say, the homology of three points. How exciting.
\begin{lemma}
If $X$ is path-connected and non-empty, then $H_0(X) \cong \Z$.
\end{lemma}
\begin{proof}
Define a homomorphism $\varepsilon: C_0(X) \to \Z$ given by
\[
\sum n_\sigma \sigma \mapsto \sum n_\sigma.
\]
Then this is surjective. We claim that the composition
\[
\begin{tikzcd}
C_1(X) \ar[r, "d"] & C_0(X) \ar[r, "\varepsilon"] & \Z
\end{tikzcd}
\]
is zero. Indeed, each simplex has two ends, and a $\sigma: \Delta^1 \to X$ is mapped to $\sigma \circ \delta_0 - \sigma \circ \delta_1$, which is mapped by $\varepsilon$ to $1 - 1 = 0$.
Thus, we know that $\varepsilon(\sigma) = \varepsilon (\sigma + d \tau)$ for any $\sigma \in C_0(X)$ and $\tau \in C_1(X)$. So we obtain a well-defined map $\varepsilon: H_0(X) \to \Z$ mapping $[x] \mapsto \varepsilon(x)$, and this is surjective as $X$ is non-empty.
So far, this is true for a general space. Now we use the path-connectedness condition to show that this map is indeed injective. Suppose $\sum n_\sigma \sigma \in C_0(X)$ lies in $\ker \varepsilon$. We choose an $x_0 \in X$. As $X$ is path-connected, for each of $\Delta^0 \to X$ we can choose a path $\tau_\sigma: \Delta^1 \to X$ with $\tau_\sigma \circ \delta_0 = \sigma$ and $\tau_\sigma \circ \delta_1 = x_0$.
Given these $1$-simplices, we can form a $1$-chain $\sum n_\sigma \tau_\sigma \in C_1(X)$, and
\[
d_1\left(\sum n_\sigma \tau_\sigma\right)= \sum n_\sigma(\sigma + x_0) = \sum n_\sigma \cdot \sigma - \left(\sum n_\sigma\right) x_0.
\]
Now we use the fact that $\sum n_\sigma = 0$. So $\sum n_\sigma \cdot \sigma$ is a boundary. So it is zero in $H_0(X)$.
\end{proof}
Combining with the coproduct formula, we have
\begin{prop}
For any space $X$, we have $H_0(X)$ is a free abelian group generated by the path components of $X$.
\end{prop}
These are pretty much the things we can do by hand. To do more things, we need to use some tools.
\section{Four major tools of (co)homology}
We now state four major properties of cohomology. At the end, we will use these properties to compute a lot of homology groups. The proofs are long and boring, so we postpone them until we have gone through the properties and enjoyed some applications of them.
\subsection{Homotopy invariance}
The first result is pretty easy to state, but it is a really powerful result that really drives what we are doing:
\begin{thm}[Homotopy invariance theorem]\index{homotopy invariance theorem}
Let $f \simeq g: X \to Y$ be homotopic maps. Then they induce the same maps on (co)homology, i.e.
\[
f_* = g_*: H_\Cdot(X) \to H_\Cdot(Y)
\]
and
\[
f^* = g^* : H^\Cdot(Y) \to H^\Cdot(X).
\]
\end{thm}
\begin{cor}
If $f: X \to Y$ is a homotopy equivalence, then $f_*: H_{\Cdot}(X) \to H_{\Cdot}(Y)$ and $f^*: H^{\Cdot}(Y) \to H^{\Cdot}(X)$ are isomorphisms.
\end{cor}
\begin{proof}
If $g: Y \to X$ is a homotopy inverse, then
\[
g_* \circ f_* = (g \circ f)_* = (\id_X)_* = \id_{H_{\Cdot}(X)}.
\]
Similarly, we have $f_* \circ g_* = (\id_Y)_* = \id_{H_{\Cdot}(Y)}$. So $f_*$ is an isomorphism with an inverse $g_*$.
The case for cohomology is similar.
\end{proof}
Since we know that $\R^n$ is homotopy equivalent to a point, it immediately follows that:
\begin{eg}
We have
\[
H^n(\R^n) = H_n(\R^n) =
\begin{cases}
\Z & n = 0\\
0 & n > 0
\end{cases}.
\]
\end{eg}
\subsection{Mayer-Vietoris}
The next tool is something that allows us to compute (co)homology by computing the (co)homology of smaller subspaces. Unfortunately, the result doesn't just say we can directly add the cohomologies, or anything like that. Instead, the information is encoded in an \emph{exact sequence}.
\begin{defi}[Exact sequence]\index{exact sequence}
We say a pair of homomorphisms
\[
\begin{tikzcd}
A \ar[r, "f"] & B \ar[r, "g"] & C
\end{tikzcd}
\]
is \emph{exact at $B$} if $\im(f) = \ker(g)$.
We say a sequence
\[
\begin{tikzcd}
\cdots \ar[r] & X_0 \ar[r] & X_1 \ar[r] & X_2 \ar[r] & \cdots
\end{tikzcd}
\]
is \emph{exact} if it is exact at each $X_n$.
\end{defi}
\begin{eg}
If
\[
\begin{tikzcd}
0 \ar[r] & A \ar[r, "f"] & B \ar[r] & 0
\end{tikzcd}.
\]
is exact, then $f$ is an isomorphism.
\end{eg}
\begin{defi}[Short exact sequence]\index{short exact sequence}
A \emph{short exact sequence} is a sequence of the form
\[
\begin{tikzcd}
0 \ar[r] & A \ar[r] & B \ar[r] & C \ar[r] & 0
\end{tikzcd}.
\]
\end{defi}
It is an easy consequence of the first isomorphism theorem that
\begin{lemma}
In a short exact sequence
\[
\begin{tikzcd}
0 \ar[r] & A \ar[r, "f"] & B \ar[r, "g"] & C \ar[r] & 0
\end{tikzcd},
\]
the map $f$ is injective; $g$ is surjective, and $C \cong B/A$.
\end{lemma}
\begin{eg}
Consider the sequence
\[
\begin{tikzcd}
0 \ar[r] & \Z/n\Z \ar[r] & A \ar[r] & \Z/m\Z \ar[r] & 0
\end{tikzcd}
\]
There are many possible values of $A$. A few obvious examples are $A = \Z/nm\Z$ and $A = \Z/n\Z \oplus \Z/m\Z$. If $n$ and $m$ are not coprime, then these are different groups.
\end{eg}
Many of the important theorems for computing homology groups tell us that the homology groups of certain spaces fall into some exact sequence, and we have to try figure out what the groups are.
\begin{thm}[Mayer-Vietoris theorem]\index{Mayer-Vietoris theorem}
Let $X = A \cup B$ be the union of two open subsets. We have inclusions
\[
\begin{tikzcd}
A \cap B \ar[r, "i_A", hook] \ar[d, "i_B", hook] & A \ar[d, hook, "j_A"]\\
B \ar[r, hook, "j_B"] & X
\end{tikzcd}.
\]
Then there are homomorphisms $\partial_{MV}: H_n(X) \to H_{n - 1}(A \cap B)$ such that the following sequence is exact:
\[
\begin{tikzcd}
\ar[r, "\partial_{MV}"] & H_n(A \cap B) \ar[r, "i_{A*} \oplus i_{B*}"] & H_n(A) \oplus H_n(B) \ar[r, "j_{A*} - j_{B*}"] & H_n(X)\ar[out=0, in=180, looseness=2, overlay, lld, "\partial_{MV}"']\\
& H_{n - 1}(A \cap B) \ar[r, "i_{A*} \oplus i_{B*}"] & H_{n - 1}(A) \oplus H_{n - 1}(B) \ar[r, "j_{A*} - j_{B*}"] & H_{n - 1}(X) \ar [r] & \cdots\\
&\cdots \ar[r] & H_0(A) \oplus H_0(B) \ar[r, "j_{A*} - j_{B*}"] & H_0(X) \ar [r] & 0
\end{tikzcd}
\]
Furthermore, the Mayer-Vietoris sequence is \emph{natural}, i.e.\ if $f: X = A\cup B \to Y = U \cup V$ satisfies $f(A) \subseteq U$ and $f(B) \subseteq V$, then the diagram
\[
\begin{tikzcd}
H_{n + 1}(X) \ar[r, "\partial_{MV}"] \ar[d, "f_*"] & H_{n}(A \cap B) \ar[r, "i_{A*} \oplus i_{B*}"] \ar[d, "f|_{A \cap B*}"] & H_n(A) \oplus H_n(B) \ar[r, "j_{A*} - j_{B*}"] \ar[d, "f|_{A*} \oplus f|_{B*}"] & H_n(X) \ar[d, "f_*"]\\
H_{n + 1}(Y) \ar[r, "\partial_{MV}"] & H_{n}(U \cap V) \ar[r, "i_{U*} \oplus i_{V*}"] & H_n(U) \oplus H_n(V) \ar[r, "j_{U*} - j_{V*}"] & H_n(Y)
\end{tikzcd}
\]
commutes.
\end{thm}
For certain elements of $H_n(X)$, we can easily specify what $\partial_{MV}$ does to it. The meat of the proof is to show that every element of $H_n(X)$ can be made to look like that. If $[a + b] \in H_n(X)$ is such that $a \in C_n(A)$ and $b \in C_n(B)$, then the map $\partial_{MV}$ is specified by
\[
\partial_{MV}([a + b]) = [d_n(a)] = [-d_n(b)] \in H_{n - 1}(A \cap B).
\]
To see this makes sense, note that we have $d_n(a + b) = 0$. So $d_n(a) = - d_n(b)$. Moreover, since $a \in C_n(A)$, we have $d_n(a) \in C_{n - 1}(A)$. Similarly, $d_n(b) \in C_{n - 1}(B)$. So $d_n(a) = - d_n(b) \in C_{n - 1}(A) \cap C_{n - 1}(B) = C_{n - 1}(A \cap B)$.
\subsection{Relative homology}
The next result again gives us another exact sequence. This time, the exact sequence involves another quantity known as \emph{relative homology}.
\begin{defi}[Relative homology]\index{relative homology}
Let $A \subseteq X$. We write $i: A \to X$ for the inclusion map. Then the map $i_n: C_n(A) \to C_n(X)$ is injective as well, and we write
\[
C_n(X, A) = \frac{C_n(X)}{C_n(A)}.
\]
The differential $d_n: C_n(X) \to C_{n - 1}(X)$ restricts to a map $C_n(A) \to C_{n - 1}(A)$, and thus gives a well-defined differential $d_n: C_n(X, A) \to C_{n - 1}(X, A)$, sending $[c] \mapsto [d_n(c)]$. The \emph{relative homology} is given by
\[
H_n(X, A) = H_n(C_{\Cdot}(X, A)).
\]
\end{defi}
We think of this as chains in $X$ where we ignore everything that happens in $A$.
\begin{thm}[Exact sequence for relative homology]\index{exact sequence for relative homology}
There are homomorphisms $\partial: H_n(X, A) \to H_{n - 1}(A)$ given by mapping
\[
\big[[c]\big] \mapsto [d_n c].
\]
This makes sense because if $c \in C_n(X)$, then $[c] \in C_n(X)/C_n(A)$. We know $[d_n c] = 0 \in C_{n - 1}(X)/C_{n - 1}(A)$. So $d_n c \in C_{n - 1}(A)$. So this notation makes sense.
Moreover, there is a long exact sequence
\[
\begin{tikzcd}
\cdots \ar[r, "\partial"] & H_n(A) \ar[r, "i_*"] & H_n(X) \ar[r, "q_*"] & H_n(X, A)\ar[out=0, in=180, looseness=2, overlay, lld, "\partial"']\\
& H_{n - 1}(A) \ar[r, "i_*"] & H_{n - 1}(X) \ar[r, "q_*"] & H_{n - 1}(X, A) \ar [r] & \cdots\\
&\cdots \ar[r] & H_0(X) \ar[r, "q_*"] & H_0(X, A) \ar [r] & 0
\end{tikzcd},
\]
where $i_*$ is induced by $i: C_{\Cdot}(A) \to C_{\Cdot}(X)$ and $q_*$ is induced by the quotient $q: C_{\Cdot}(X) \to C_{\Cdot}(X, A)$.
\end{thm}
This, again, is natural. To specify the naturality condition, we need the following definition:
\begin{defi}[Map of pairs]\index{map of pairs}
Let $(X, A)$ and $(Y, B)$ be topological spaces with $A \subseteq X$ and $B \subseteq Y$. A \emph{map of pairs} is a map $f: X \to Y$ such that $f(A) \subseteq B$.
\end{defi}
Such a map induces a map $f_*: H_n(X, A) \to H_n(Y, B)$, and the exact sequence for relative homology is natural for such maps.
\subsection{Excision theorem}
Now the previous result is absolutely useless, because we just introduced a new quantity $H_n(X, A)$ we have no idea how to compute again. The main point of relative homology is that we want to think of $H_n(X, A)$ as the homology of $X$ when we ignore $A$. Thus, one might expect that the relative homology does not depend on the things ``inside $A$''. However, it is not true in general that, say, $H_n(X, A) = H_n(X \setminus A)$. Instead, what we are allowed to do is to remove subspaces of $A$ that are ``not too big''. This is given by excision:
\begin{thm}[Excision theorem]\index{excision theorem}
Let $(X, A)$ be a pair of spaces, and $Z \subseteq A$ be such that $\overline{Z} \subseteq \mathring{A}$ (the closure is taken in $X$). Then the map
\[
H_n(X \setminus Z, A \setminus Z) \to H_n(X, A)
\]
is an isomorphism.
\end{thm}
\begin{center}
\begin{tikzpicture}
\draw [draw=black, fill=morange, opacity=0.5] (0, 0) rectangle (3, 2);
\node at (0.5, 0.5) {$X$};
\draw [draw=black, fill=mblue, opacity=0.6] (1.8, 1) circle [radius=0.7];
\node at (2.6, 1.7) {$A$};
\draw [draw=black, fill=mred, opacity=0.8, decorate, decoration={snake}] (1.75, 0.95) circle [radius=0.4] node [white] {$Z$};
\end{tikzpicture}
\end{center}
While we've only been talking about homology, everything so far holds analogously for cohomology too. It is again homotopy invariant, and there is a Mayer-Vietoris sequence (with maps $\partial_{MV}: H^n(A \cap B) \to H^{n + 1}(X)$). The relative cohomology is defined by $C^{\Cdot}(X, A) = \Hom(C_{\Cdot}(X, A), \Z)$ and so $H^*(X, A)$ is the cohomology of that. Similarly, excision holds.
\subsection{Applications}
We now use all these tools to do lots of computations. The first thing we compute will be the homology of spheres.
\begin{thm}
We have
\[
H_i(S^1) =
\begin{cases}
\Z & i = 0, 1\\
0 & \text{otherwise}
\end{cases}.
\]
\end{thm}
\begin{proof}
We can split $S^1$ up as
\begin{center}
\begin{tikzpicture}
\draw circle [radius=1];
\draw [red] (1.105, -0.4673) arc(-22.9:202.9:1.2);
\node [red, above] at (0, 1.2) {$A$};
\draw [blue] (1.279, 0.545) arc(22.9:-202.9:1.4);
\node [blue, below] at (0, -1.4) {$B$};
\node [circ] at (1, 0) {};
\node at (1, 0) [left] {$q$};
\node [circ] at (-1, 0) {};
\node at (-1, 0) [right] {$p$};
\end{tikzpicture}
\end{center}
We want to apply Mayer-Vietoris. We have
\[
A \cong B \cong \R \simeq *,\quad A \cap B \cong \R \coprod \R \simeq \{p\} \coprod \{q\}.
\]
We obtain
\[
\begin{tikzcd}[row sep=tiny]
& 0 \ar[d, equals] & 0 \ar[d, equals] \\
\cdots \ar[r] & H_1(A \cap B) \ar[r] & H_1(A) \oplus H_1(B) \ar[r] & H_1(S^1)\ar[out=0, in=180, looseness=2, overlay, lldd, "\partial"']\\
\vphantom{a}\\
& H_0(A \cap B) \ar[r, "i_{A*} \oplus i_{B*}"] & H_0(A) \oplus H_0(B) \ar[r] & H_0(S^1) \ar [r] & 0\\
& \Z \oplus \Z \ar[u, equals] & \Z \oplus Z \ar[u, equals] & \Z \ar[u, equals]
\end{tikzcd}
\]
Notice that the map into $H_1(S^1)$ is zero. So the kernel of $\partial$ is trivial, i.e.\ $\partial$ is an injection. So $H_1(S^1)$ is isomorphic to the image of $\partial$, which is, by exactness, the kernel of $i_{A*} \oplus i_{B*}$. So we want to know what this map does.
We know that $H_0(A \cap B) \cong \Z \oplus \Z$ is generated by $p$ and $q$, and the inclusion map sends each of $p$ and $q$ to the unique connected components of $A$ and $B$. So the homology classes are both sent to $(1, 1) \in H_0(A) \oplus H_0(B) \cong \Z \oplus \Z$. We then see that the kernel of $i_{A*} \oplus i_{B*}$ is generated by $(p - q)$, and is thus isomorphic to $\Z$. So $H_1(S^1) \cong \Z$.
By looking higher up the exact sequence, we see that all other homology groups vanish.
\end{proof}
We can do the sphere in general in the same way.
\begin{thm}
For any $n \geq 1$, we have
\[
H_i(S^n) =
\begin{cases}
\Z & i = 0, n\\
0 & \text{otherwise}
\end{cases}.
\]
\end{thm}
\begin{proof}
We again cut up $S^n$ as
\begin{align*}
A &= S^n \setminus \{N\} \cong \R^n \simeq *,\\
B &= S^n \setminus \{S\} \cong \R^n \simeq *,
\end{align*}
where $N$ and $S$ are the north and south poles. Moreover, we have
\[
A\cap B \cong \R \times S^{n - 1} \simeq S^{n - 1}
\]
So we can ``induct up'' using the Mayer-Vietoris sequence:
\[
\begin{tikzcd}
\cdots \ar[r] & H_i(S^{n - 1}) \ar[r] & H_i(*) \oplus H_i(*) \ar[r] & H_i(S^n)\ar[out=0, in=180, looseness=2, overlay, lld, "\partial"]\\
& H_{i - 1}(S^{n - 1}) \ar[r] & H_{i - 1}(*) \oplus H_{i - 1}(*) \ar[r] & H_{i - 1}(S^n) \ar [r] & \cdots
\end{tikzcd}
\]
Now suppose $n \geq 2$, as we already did $S^1$ already. If $i > 1$, then $H_i(*) = 0 = H_{i - 1}(*)$. So the Mayer-Vietoris map
\[
\begin{tikzcd}
H_i(S^n) \ar[r, "\partial"] & H_{i - 1}(S^{n - 1})
\end{tikzcd}
\]
is an isomorphism.
All that remains is to look at $i = 0, 1$. The $i = 0$ case is trivial. For $i = 1$, we look at
\[
\begin{tikzcd}[row sep=tiny]
& & 0 \ar[d, equals] \\
& \cdots \ar[r] & H_1(*) \oplus H_1(*) \ar[r] & H_1(S^n)\ar[out=0, in=180, looseness=2, overlay, lldd, "\partial"]\\
\vphantom{a} \\
& H_0(S^{n - 1}) \ar[r, "f"] & H_0(*) \oplus H_0(*) \ar[r] & H_0(S^n) \ar [r] & 0\\
& \Z \ar[u, equals] & \Z \oplus \Z \ar[u, equals] & \Z \ar[u, equals]
\end{tikzcd}
\]
To conclude that $H_1(S^n)$ is trivial, it suffices to show that the map $f$ is injective. By picking the canonical generators, it is given by $1 \mapsto (1, 1)$. So we are done.
\end{proof}
\begin{cor}
If $n \not= m$, then $S^{n - 1} \not\simeq S^{m - 1}$, since they have different homology groups.
\end{cor}
\begin{cor}
If $n \not= m$, then $\R^n \not \cong \R^m$.
\end{cor}
Now suppose we have a map $f: S^n \to S^n$. As always, it induces a map $f_*: H_n(S^n) \to H_n(S^n)$. Since $H_n (S^n) \cong \Z$, we know the map $f_*$ is given by multiplication by some integer. We call this the \emph{degree} of the map $f$.
\begin{defi}[Degree of a map]\index{degree of map}
Let $f: S^n \to S^n$ be a map. The \emph{degree} $\deg(f)$ is the unique integer such that under the identification $H_n(S^n) \cong \Z$, the map $f_*$ is given by multiplication by $\deg(f)$.
\end{defi}
In particular, for $n = 1$, the degree is the winding number.
Note that there are two ways we can identify $H_n(S^n)$ and $\Z$, which differ by a sign. For the degree to be well-defined, and not just defined up to a sign, we must ensure that we identify both $H_n(S^n)$ in the same way.
Using the degree, we can show that certain maps are \emph{not} homotopic to each other. We first note the following elementary properties:
\begin{prop}\leavevmode
\begin{enumerate}
\item $\deg(\id_{S^n}) = 1$.
\item If $f$ is not surjective, then $\deg(f) = 0$.
\item We have $\deg(f\circ g) = (\deg f)(\deg g)$.
\item Homotopic maps have equal degrees.
\end{enumerate}
\end{prop}
\begin{proof}\leavevmode
\begin{enumerate}
\item Obvious.
\item If $f$ is not surjective, then $f$ can be factored as
\[
\begin{tikzcd}
S^n \ar[r, "f"] & S^n \setminus \{p\} \ar[r, hook] & S^n
\end{tikzcd},
\]
where $p$ is some point not in the image of $f$. But $S^n \setminus \{p\}$ is contractible. So $f_*$ factors as
\[
\begin{tikzcd}
f_*: H_n(S^n) \ar[r] & H_n(*) = 0 \ar[r] & H_n(S^n)
\end{tikzcd}.
\]
So $f_*$ is the zero homomorphism, and is thus multiplication by $0$.
\item This follows from the functoriality of $H_n$.
\item Obvious as well.\qedhere
\end{enumerate}
\end{proof}
As a corollary, we obtain the renowned Brouwer's fixed point theorem, a highly non-constructive fixed point existence theorem proved by a constructivist.
\begin{cor}[Brouwer's fixed point theorem]\index{Brouwer's fixed point theorem}
Any map $f: D^n \to D^n$ has a fixed point.\index{fixed point}
\end{cor}
\begin{proof}
Suppose $f$ has no fixed point. Define $r: D^n \to S^{n - 1} = \partial D^n$ by taking the intersection of the ray from $f(x)$ through $x$ with $\partial D^n$. This is continuous.
\begin{center}
\begin{tikzpicture}
\draw circle [radius=2cm];
\node [circ] at (-0.4, -0.3) {};
\node at (-0.4, -0.3) [below] {$x$};
\node [circ] at (0.4, 0.3) {};
\node at (0.4, 0.3) [right] {$f(x)$};
\draw (0.4, 0.3) -- (-1.6, -1.2);
\node at (-1.6, -1.2) [circ] {};
\node at (-1.6, -1.2) [anchor = north east] {$r(x)$};
\end{tikzpicture}
\end{center}
Now if $x \in \partial D^n$, then $r(x) = x$. So we have a map
\[
\begin{tikzcd}
S^{n - 1} = \partial D^n \ar[r, "i"] & D^n \ar[r, "r"] & \partial D^n = S^{n - 1}
\end{tikzcd},
\]
and the composition is the identity. This is a contradiction --- contracting $D^n$ to a point, this gives a homotopy from the identity map $S^{n - 1} \to S^{n - 1}$ to the constant map at a point. This is impossible, as the two maps have different degrees.
\end{proof}
A more manual argument to show this would be to apply $H_{n - 1}$ to the chain of maps above to obtain a contradiction.
We know there is a map of degree $1$, namely the identity, and a map of degree $0$, namely the constant map. How about other degrees?
\begin{prop}
A reflection $r: S^n \to S^n$ about a hyperplane has degree $-1$. As before, we cover $S^n$ by
\begin{align*}
A &= S^n \setminus \{N\} \cong \R^n \simeq *,\\
B &= S^n \setminus \{S\} \cong \R^n \simeq *,
\end{align*}
where we suppose the north and south poles lie in the hyperplane of reflection. Then both $A$ and $B$ are invariant under the reflection. Consider the diagram
\[
\begin{tikzcd}
H_n(S^n) \ar[r, "\partial_{MV}", "\sim"'] \ar[d, "r_*"] & H_{n - 1}(A \cap B) \ar[d, "r_*"] & H_{n - 1}(S^{n - 1}) \ar[l, "\sim"] \ar[d, "r_*"]\\
H_n(S^n) \ar[r, "\partial_{MV}", "\sim"'] & H_{n - 1}(A \cap B) & H_{n - 1}(S^{n - 1}) \ar[l, "\sim"]
\end{tikzcd}
\]
where the $S^{n - 1}$ on the right most column is given by contracting $A \cap B$ to the equator. Note that $r$ restricts to a reflection on the equator. By tracing through the isomorphisms, we see that $\deg(r) = \deg(r|_{\mathrm{equator}})$. So by induction, we only have to consider the case when $n = 1$. Then we have maps
\[
\begin{tikzcd}
0 \ar[r] & H_1(S^1) \ar[r, "\partial_{MV}", "\sim"'] \ar[d, "r_*"] & H_0(A \cap B) \ar[d, "r_*"] \ar[r] & H_0(A) \oplus H_0(B) \ar[d, "r_* \oplus r_*"]\\
0 \ar[r] & H_1(S^1) \ar[r, "\partial_{MV}", "\sim"'] & H_0(A \cap B) \ar[r] & H_0(A) \oplus H_0(B)
\end{tikzcd}
\]
Now the middle vertical map sends $p \mapsto q$ and $q \mapsto p$. Since $H_1(S^1)$ is given by the kernel of $H_0(A \cap B) \to H_0(A) \oplus H_0(B)$, and is generated by $p - q$, we see that this sends the generator to its negation. So this is given by multiplication by $-1$. So the degree is $-1$.
\end{prop}
\begin{cor}
The antipodal map $a: S^n \to S^n$ given by
\[
a(x_1, \cdots, x_{n + 1}) = (-x_1, \cdots, -x_{n + 1})
\]
has degree $(-1)^{n + 1}$ because it is a composition of $(n + 1)$ reflections.
\end{cor}
\begin{cor}[Hairy ball theorem]\index{Hairy ball theorem}
$S^n$ has a nowhere $0$ vector field iff $n$ is odd. More precisely, viewing $S^n \subseteq \R^{n + 1}$, a vector field on $S^n$ is a map $v: S^n \to \R^{n + 1}$ such that $\bra v(x), x\ket = 0$, i.e.\ $v(x)$ is perpendicular to $x$.
\end{cor}
\begin{proof}
If $n$ is odd, say $n = 2k - 1$, then
\[
v(x_1, y_1, x_2, y_2, \cdots, x_k, y_k) = (y_1, -x_1, y_2, -x_2, \cdots, y_k, -x_k)
\]
works.
Conversely, if $v: S^n \to \R^{n + 1} \setminus \{0\}$ is a vector field, we let $w = \frac{v}{\abs{v}}: S^n \to S^n$. We can construct a homotopy from $w$ to the antipodal map by ``linear interpolation'', but in a way so that it stays on the sphere. We let
\begin{align*}
H: [0, \pi] \times S^n &\to S^n\\
(t, x) &\mapsto \cos(t) x + \sin(t) w(x)
\end{align*}
Since $w(x)$ and $x$ are perpendicular, it follows that this always has norm $1$, so indeed it stays on the sphere.
Now we have
\[
H(0, x) = x,\quad H(\pi, x) = -x.
\]
So this gives a homotopy from $\id$ to $a$, which is a contradiction since they have different degrees.
\end{proof}
So far, we've only been talking about spheres all the time. We now move on to something different.
\begin{eg}
Let $K$ be the Klein bottle. We cut them up by
\begin{center}
\begin{tikzpicture}
\draw [->-=0.55, mred] (0, 0) -- (3, 0);
\draw [->-=0.55, mred] (3, 3) -- (0, 3);
\draw [->-=0.55, mblue] (3, 0) -- (3, 3);
\draw [->-=0.55, mblue] (0, 0) -- (0, 3);
\fill [morange, opacity=0.5] (0, 0) rectangle (3, 1.2);
\fill [morange, opacity=0.5] (0, 1.8) rectangle (3, 3);
\node [right, morange] at (3, 2.5) {$A$};
\fill [mgreen, opacity=0.5] (0, 0.8) rectangle (3, 2.2);
\node [right, mgreen] at (3, 1.5) {$B$};
\foreach \n in {0,1,2,3,4,5,6,7,8,9,10,11,12,13} {
\begin{scope}[shift={(0.15 + 0.2 * \n, 0.9)}]
\fill (0, 0) -- (0.05, 0) -- (0.1, 0.1) -- (0.05, 0.2) -- (0, 0.2) -- (0.05, 0.1) -- cycle;