\classheader{2012-11-26}

%(coming soon)

Today we'll be talking about the fundamental group. It may seem like it would be similar to homology, but in homology, we only deal with abelian / linear things, whereas the fundamental group is usually non-abelian.

\subsection*{The Fundamental Group}

We start by considering maps from $[0,1]$ to $X$. If $\alpha,\beta:[0,1]\to X$ satisfy $\alpha(1)=\beta(0)$, then we can define their composition $(a\ast b):[0,1]\to X$ by
\[(a\ast b)(t)=\begin{cases}
\alpha(2t)&\text{if }0\leq t\leq \frac{1}{2},\\
\beta(2t-1)&\text{if }\frac{1}{2}\leq t\leq 1.
\end{cases}\]
We define a ``path'' to be a map from $[0,1]$ to $X$, with the endpoints treated specially. We define two paths to be homotopic when they are homotopic rel their endpoints. 

Thus, if $\alpha$ and $\beta$ are paths with $\alpha(0)=\beta(0)$ and $\alpha(1)=\beta(1)$, then we say $\alpha\sim\beta$ when there is some homotopy $F:[0,1]^2\to X$ such that $F_0=\alpha$, $F_1=\beta$, $F_t(0)=\alpha(0)=\beta(0)$, and $F_t(1)=\alpha(1)=\beta(1)$.

Now we'll prove this is an equivalence relation.

For any path $a$, we have $a\sim a$ because we can define a homotopy as follows (imagine time progresses as one goes bottom to top; this homotopy is constant on the gray lines):

\begin{center}
\begin{tikzpicture}[scale=2]
\foreach \x in {0.1,0.2,...,0.9} {
\draw[gray] (\x,0) -- (\x,1);}
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$\alpha$};
\node at (0.5,-0.2) {$\alpha$};
\node at (-0.3,0.5) {$\alpha(0)$};
\node at (1.3,0.5) {$\alpha(1)$};
\end{tikzpicture}
\end{center}

Suppose $\alpha$ and $\beta$ are paths with $\alpha(0)=\beta(0)$ and $\alpha(1)=\beta(1)$, such that $\alpha\sim \beta$ (say, via a path homotopy $F:[0,1]^2\to X$ from $F_0=\alpha$ to $F_1=\beta$). Then we have $\beta\sim \alpha$, because we can define a path homotopy $G:[0,1]^2\to X$ from $G_0=\beta$ to $G_1=\alpha$ simply as $G_t(s)=F_{1-t}(s)$. Pictorally, (again, imagine time progresses as one goes from bottom to top):

\begin{center}
\begin{tikzpicture}[scale=2]
\draw[thick,decorate,decoration={brace, amplitude=4pt}] 
        (-0.6,0) -- (-0.6,1) node[midway, left=3pt]{$F$};
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$\beta$};
\node at (0.5,-0.2) {$\alpha$};
\node at (-0.3,0.2) {$\alpha(0)$};
\node at (1.3,0.2) {$\alpha(1)$};
\node[rotate=90] at (-0.3,0.5) {$=$};
\node[rotate=90] at (1.3,0.5) {$=$};
\node at (-0.3,0.8) {$\beta(0)$};
\node at (1.3,0.8) {$\beta(1)$};
\node at (0.5,0.5) {$F$};
\end{tikzpicture}\qquad\qquad\qquad\begin{tikzpicture}[scale=2]
\draw[thick,decorate,decoration={brace, amplitude=4pt}] 
        (-0.6,0) -- (-0.6,1) node[midway, left=3pt]{$G$};
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$\alpha$};
\node at (0.5,-0.2) {$\beta$};
\node at (-0.3,0.2) {$\beta(0)$};
\node at (1.3,0.2) {$\beta(1)$};
\node[rotate=90] at (-0.3,0.5) {$=$};
\node[rotate=90] at (1.3,0.5) {$=$};
\node at (-0.3,0.8) {$\alpha(0)$};
\node at (1.3,0.8) {$\alpha(1)$};
\node[yscale=-1] at (0.5,0.5) {$F$};
\end{tikzpicture}
\end{center}

Lastly, if $\alpha,\beta,\gamma$ are paths with $\alpha(0)=\beta(0)=\gamma(0)$ and $\alpha(1)=\beta(1)=\gamma(1)$, such that $\alpha\sim\beta$ and $\beta\sim \gamma$ via homotopies $F:[0,1]^2\to X$ from $F_0=\alpha$ to $F_1=\beta$ and $G:[0,1]^2\to X$ from $G_0=\beta$ to $G_1=\gamma$, we have $\alpha\sim \gamma$ because we can define a homotopy $H:[0,1]^2\to X$ from $H_0=\alpha$ to $H_1=\gamma$ by
\[H_t(s)=\begin{cases}
F_{2t}(s) & \text{ if }0\leq t\leq\frac{1}{2},\\
G_{2t-1}(s) & \text{ if }\frac{1}{2}\leq t\leq 1.
\end{cases}\]
Pictorally,
\begin{center}
\begin{tikzpicture}[scale=2]
\draw[thick,decorate,decoration={brace, amplitude=4pt}] 
        (-0.6,0) -- (-0.6,1) node[midway, left=3pt]{$F$};
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$\beta$};
\node at (0.5,-0.2) {$\alpha$};
\node at (-0.3,0.2) {$\alpha(0)$};
\node at (1.3,0.2) {$\alpha(1)$};
\node[rotate=90] at (-0.3,0.5) {$=$};
\node[rotate=90] at (1.3,0.5) {$=$};
\node at (-0.3,0.8) {$\beta(0)$};
\node at (1.3,0.8) {$\beta(1)$};
\node at (0.5,0.5) {$F$};
\end{tikzpicture}\qquad\qquad\begin{tikzpicture}[scale=2]
\draw[thick,decorate,decoration={brace, amplitude=4pt}] 
        (-0.6,0) -- (-0.6,1) node[midway, left=3pt]{$G$};
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$\beta$};
\node at (0.5,-0.2) {$\gamma$};
\node at (-0.3,0.2) {$\gamma(0)$};
\node at (1.3,0.2) {$\gamma(1)$};
\node[rotate=90] at (-0.3,0.5) {$=$};
\node[rotate=90] at (1.3,0.5) {$=$};
\node at (-0.3,0.8) {$\beta(0)$};
\node at (1.3,0.8) {$\beta(1)$};
\node at (0.5,0.5) {$G$};
\end{tikzpicture}\\
\begin{tikzpicture}[scale=2.5]
\draw[thick,decorate,decoration={brace, amplitude=4pt}] 
        (-0.6,0) -- (-0.6,1) node[midway, left=3pt]{$H$};
\draw[thick] (0,0) rectangle (1,1);
\draw[thick] (0,0.5) -- node [pos = 0.65,auto,shift={(0.0,-0.07)}] {$\beta$} (1,0.5);
\node at (0.65,1.15) {$\gamma$};
\node at (0.65,-0.15) {$\alpha$};
\node at (-0.3,0.2) {$\alpha(0)$};
\node at (1.3,0.2) {$\alpha(1)$};
\node[rotate=90] at (-0.3,0.5) {$=$};
\node[rotate=90] at (1.3,0.5) {$=$};
\node at (-0.3,0.8) {$\gamma(0)$};
\node at (1.3,0.8) {$\gamma(1)$};
\node at (0.5,0.25) {$F$};
\node at (0.5,0.75) {$G$};
\end{tikzpicture}
\end{center}

Note also that if $\alpha\ast\alpha'$ and $\beta\sim\beta'$, then we also have $(\alpha\ast\beta)\sim (\alpha'\ast\beta')$:
\begin{center}
\begin{tikzpicture}[scale=2.5]
\draw[thick] (0,0) rectangle (1,1);
\draw[thick] (0.5,0) -- (0.5,1);
\node at (0.25,1.15) {$\alpha'$};
\node at (0.75,1.15) {$\beta'$};
\node at (0.25,-0.15) {$\alpha$};
\node at (0.75,-0.15) {$\beta$};
\node at (-0.3,0.2) {$\alpha(0)$};
\node at (1.3,0.2) {$\beta(1)$};
\node[rotate=90] at (-0.3,0.5) {$=$};
\node[rotate=90] at (1.3,0.5) {$=$};
\node at (-0.3,0.8) {$\alpha'(0)$};
\node at (1.3,0.8) {$\beta'(1)$};
\node at (0.25,0.5) {$F$};
\node at (0.75,0.5) {$G$};
\end{tikzpicture}
\end{center}

If $[\alpha]$ denotes the equivalence class of $\alpha$ with respect to $\sim$, then this demonstrates that $\ast$ descends to an operation on equivalence classes, i.e. $[\alpha\ast\beta]=[\alpha]\ast[\beta]$.

Moreover, $\ast$ is associative (when $\ast$ is defined) because there is a homotopy
\begin{center}
\begin{tikzpicture}[scale=2.5]
\draw[thick] (0,0) rectangle (1,1);
\draw[thick] (0.5,0) -- (0.25,1);
\draw[thick] (0.75,0) -- (0.5,1);
\node at (0.25,-0.15) {$\alpha$};
\node at (0.625,-0.15) {$\beta$};
\node at (0.875,-0.15) {$\gamma$};
\node at (0.125,1.15) {$\alpha$};
\node at (0.375,1.15) {$\beta$};
\node at (0.75,1.15) {$\gamma$};
\node at (-0.3,0.5) {$\alpha(0)$};
\node at (1.3,0.5) {$\gamma(1)$};
\node at (0.5,-0.4) {$\alpha\ast(\beta\ast\gamma)$};
\node at (0.5,1.4) {$(\alpha\ast\beta)\ast\gamma$};
\end{tikzpicture}
\end{center}

There are also ``identities'' for $\ast$ (be careful, because $\ast$ is only defined for paths whose endpoints are the same). We define $\mathbf{1}_p$ to be the constant map from $[0,1]$ to $p\in X$. Then $\mathbf{1}_{\alpha(0)}\ast \alpha=\alpha$ because there is a homotopy
\begin{center}
\begin{tikzpicture}[scale=2.5]
\draw[thick] (0,0) rectangle (1,1);
\draw[thick] (0.5,0) -- (0,1);
\node at (0.25,-0.15) {$\mathbf{1}_{\alpha(0)}$};
\node at (0.75,-0.15) {$\alpha$};
\node at (0.5,1.15) {$\alpha$};
\node at (-0.3,0.5) {$\alpha(0)$};
\node at (1.3,0.5) {$\alpha(1)$};
\end{tikzpicture}
\end{center}
and similarly $\alpha\ast \mathbf{1}_{\alpha(1)}=\alpha$.

Lastly, there are inverses. If $\gamma$ is a path, we define $\gamma^{-1}$ to be the same path but with the opposite orientation, i.e. $\gamma^{-1}(t)=\gamma(1-t)$. Then $\gamma\ast\gamma^{-1}=\mathbf{1}_{\gamma(0)}$, because there is a homotopy
\begin{center}
\begin{tikzpicture}[scale=2.5]
\draw[thick] (0,0) rectangle (1,1);
\foreach \x in {0.05,0.1,...,0.5} {
\draw[gray] (0.5-\x,0) -- (0.5-\x,2*\x) -- (0.5+\x,2*\x) -- (0.5+\x,0);
}
\draw[thick] (0.5,0) -- (0,1);
\draw[thick] (0.5,0) -- (1,1);
\node at (0.25,-0.15) {$\gamma$};
\node at (0.75,-0.15) {$\gamma^{-1}$};
\node at (0.5,1.15) {$\mathbf{1}_{\gamma(0)}$};
\node at (-0.3,0.5) {$\gamma(0)$};
\node at (1.3,0.5) {$\gamma(0)$};
\end{tikzpicture}
\end{center}
(it is constant on the gray lines), and similarly with $\gamma^{-1}\ast\gamma=\mathbf{1}_{\gamma(1)}$.

A structure satisfying these properties is called a groupoid. It is the same as a group, except there are some elements which may not be able to be composed.

Thus, given any space $X$ gives rise to a groupoid, called the fundamental groupoid of $X$. The construction of this groupoid didn't favor any point in $X$ over any other, but if we break the symmetry a bit and choose some point $x\in X$ to be our basepoint, then we can define a group as follows: let $\pi_1(X,x)$ be the set of equivalence classes of paths $\alpha:[0,1]\to X$ which both start and end at $x$, i.e. $\alpha(0)=\alpha(1)=x$. Then $\pi_1(X,x)$ forms a group under the operation $\ast$, and the identity of $\pi_1(X,x)$ is just $\mathbf{1}_x$. (Note that we usually want to assume $X$ is connected.) We say that $\pi_1(X,x)$ is the fundamental group of $X$ with basepoint $x$.


How does $\pi_1(X,x)$ depend on our choice of $x$?

Given two points $x$ and $y$, choosing a path $\gamma$ from $x$ to $y$ determines a homomorphism from $\pi_1(X,x)$ to $\pi_1(X,y)$ by sending $[\alpha]$ to $[\gamma^{-1}\ast\alpha\ast\gamma]$. In fact, it is easy to see that it must be an isomorphism (take $\gamma^{-1}$ as a path from $y$ to $x$). If $x=y$, then this isomorphism is just the inner automorphism that is conjugation by $[\gamma]$, because $[\gamma^{-1}\ast\alpha\ast\gamma]=[\gamma^{-1}]\ast[\alpha]\ast[\gamma]$.

Thus, we often say somewhat loosely that a group is ``the'' fundamental group of a space $X$, even though the isomorphism is only determined up to inner automorphism. At any rate, the isomorphism class determined.

We also want $\pi_1$ to be functorial. Given a map $f:X\to Y$ and $\alpha:[0,1]\to X$, then $(f\circ\alpha):[0,1]\to Y$ is a path in $Y$, so we get a map $f_*$ sending paths in $X$ to paths in $Y$, which is compatible with $\sim$ because $f$ will also map any homotopy:
\begin{center}
\begin{tikzpicture}[scale=2]
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$\alpha'$};
\node at (0.5,-0.2) {$\alpha$};
\node at (-0.3,0.5) {$\alpha(0)$};
\node at (1.3,0.5) {$\alpha(1)$};
\node at (0.5,0.5) {$F$};
\draw[->,>=stealth] (1.9,0.5) to[bend left=20] node[midway,auto,label={$f$}] {} (2.9,0.5);
\begin{scope}[shift={(4.3,0)}]
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$f\circ\alpha'$};
\node at (0.5,-0.2) {$f\circ\alpha$};
\node[rotate=0] at (-0.55,0.5) {$(f\circ\alpha)(0)$};
\node[rotate=0] at (1.6,0.5) {$(f\circ\alpha)(1)$};
\node at (0.5,0.5) {$f\circ F$};
\end{scope}
\end{tikzpicture}
\end{center}
and moreover $f_*$ is compatible with $\ast$. Thus, $f_*$ induces a homomorphism of groupoids, and if we choose basepoints, we get a map $f_*:\pi_1(X,x)\to\pi_1(Y,f(x))$.

If $f$ and $f'$ are maps from $X$ to $Y$ both sending $x$ to the same point, and $f\simeq f'$ rel $x$ (i.e. there is a homotopy $F:X\times [0,1]\to Y$ such that $F_t(x)=f(x)=f'(x)$), then the induced maps $f_*:\pi_1(X,x)\to \pi_1(Y,f(x))$ and $f_*':\pi_1(X,x)\to \pi_1(Y,f(x))$ are in fact the same, because the homotopy $F$ between $f$ and $f'$ gives us, for any loop $\alpha$ in $X$ based at $x$, a homotopy
\begin{center}
\begin{tikzpicture}[scale=2]
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$f'\circ \alpha$};
\node at (0.5,-0.2) {$f\circ \alpha$};
\node at (-0.3,0.5) {$f(x)$};
\node at (1.3,0.5) {$f(x)$};
\node at (0.5,0.5) {$F\circ \alpha$};
\end{tikzpicture}
\end{center}

If $f:X\to Y$ is a homotopy equivalence, then $f_*:\pi_1(X,x)\to\pi_1(Y,f(x))$ is an isomorphism. We can see this as follows. Let $g$ be the homotopy inverse of $f$, so that there is a homotopy $F:X\times [0,1]\to X$ such that $F_t(0)=(g\circ f)$ and $F_t(1)=\id_X$. Let $\gamma$ keep track of what $F$ does to $x$, so that $\gamma:[0,1]\to X$ is a path starting at $(g\circ f)(x)$ and ending at $x$. Then
the composition
\begin{center}
\begin{tikzcd}
\pi_1(X,x) \ar{r}{f_*} & \pi_1(Y,f(x)) \ar{r}{g_*} & \pi_1(X,gf(x)) \ar{r}{\gamma_*} & \pi_1(X,x)
\end{tikzcd}
\end{center}
is the identity map. Pictorally, we get immediately from our setup a homotopy (note that the left and right edges are \textbf{not} constant here, i.e. this is not a path-homotopy)
\begin{center}
\begin{tikzpicture}[scale=2]
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$gf(\alpha)$};
\node at (0.5,-0.2) {$\alpha$};
\node at (-0.3,0.5) {$\gamma$};
\node at (1.3,0.5) {$\gamma$};
\draw[-angle 90] (0,1) -- (0,0.45);
\draw[-angle 90] (1,1) -- (1,0.45);
\node at (0.5,0.5) {$F\circ \alpha$};
\end{tikzpicture}
\end{center}
which we can then deform into a nice path-homotopy by shrinking the left and right edges onto the top, leaving the left and right sides to be mapped constantly to $\alpha(0)$:
\begin{center}
\begin{tikzpicture}[scale=3]
\draw[thick] (0,0) rectangle (1,1);
\node at (0.5,1.15) {$gf(\alpha)$};
\node[mypoint] at (0.33,1) {};
\node[mypoint] at (0.66,1) {};
\node at (0.5,-0.15) {$\alpha$};
\node at (-0.2,0.5) {$\alpha(0)$};
\node at (1.2,0.5) {$\alpha(0)$};
\draw[-angle 90] (0.5,1) -- (0.16,1);
\draw[-angle 90] (0.5,1) -- (0.83,1);
\node at (0.16,1.15) {$\gamma$};
\node at (0.83,1.15) {$\gamma$};
\end{tikzpicture}
\end{center}

In summary, for a connected space $X$ with basepoint $x$, then $\pi_1(X,x)$ is a group, consisting of the homotopy calsses of paths based at $x$ rel endpoints, with $\ast$ as the operation. Changing the basepoints gives isomorphic groups. Maps $f:X\to Y$ induce homomorphisms $f_*:\pi_1(X,x)\to\pi_1(Y,f(x))$. Homotopic maps induce the same homomorphism, and homotopy equivalences induce isomorphisms.