# Abstract Nonsense

## The Fundamental Groupoid and Group (Pt. II)

Point of Post: This is a continuation of this post.

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We first verify that this pairing satisfies some of the properties we expect out of it:

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Theorem: Let $X$ be a topological space. If $x,y\in X$, $\alpha\in \mathcal{P}(x,y)$, and $c_x,c_y$ denote the constant paths $x\to x$ and $y\to y$, then $[c_x]\cdot[\alpha]=[\alpha]\cdot[c_y]=[\alpha]$

Proof: To prove that $\alpha\cdot c_y\simeq \alpha$ define $H:[0,1]^2\to X$ by

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$H(s,t)=\begin{cases}\alpha\left(\frac{2s}{t+1}\right) & \mbox{if} \quad 0\leqslant s\leqslant \frac{t+1}{2}\\ y & \mbox{if} \quad \frac{t+1}{2}\leqslant s\leqslant 1\end{cases}$

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It is clear that $H$ is continuous by the gluing lemma, since the overlap of the domains is $\displaystyle s=\frac{t+1}{2}$ where the second case gives $y$ and the first gives

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$\displaystyle \alpha\left(\frac{\displaystyle 2\frac{t+1}{2}}{t+1}\right)=\alpha(1)=y$

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Moreover,

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$\displaystyle H(s,0)=\begin{cases}\alpha(s) & \mbox{if}\quad 0\leqslant s\leqslant \tfrac{1}{2}\\ y & \mbox{if}\quad \tfrac{1}{2}\leqslant s\leqslant 1\end{cases}=(\alpha\cdot c_y)(s)$

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and

$\displaystyle H(s,1)=\begin{cases}\alpha(s) & \mbox{if} \quad 0\leqslant s\leqslant 1\\ y & \mbox{if}\quad s=1\end{cases}=\alpha(s)$

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Thus, we may conclude that $\alpha\cdot c_y\simeq \alpha$ and thus $[\alpha]\cdot[c_y]=[\alpha\cdot c_y]=[\alpha]$.

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The other case follows similarly. $\blacksquare$

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Intuitively, we can think of this homotopy as doing $c_y$ for less, and less time until we are eventually only doing it for the instant $s=1$.

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For each path $\alpha\in\mathcal{P}(x,y)$ let $\alpha^{\leftarrow}\in\mathcal{P}(y,x)$ denote the path defined by $\alpha^{\leftarrow}(s)=\alpha(1-s)$. This is the path obtained by doing $\alpha$ in the reverse. It should seem fairly obvious that if we start at a point, make a path to another point, and then take the same path in reverse, that (up to homotopy) we just get the initial path at the point. More formally:

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Theorem: Let $x,y\in X$. Then, for any $\alpha\in \mathcal{P}(x,y)$ one has that $[\alpha]\cdot[\alpha^{\leftarrow}]=[c_x]$ and $[\alpha^{\leftarrow}]\cdot[\alpha]=[c_y]$.

Proof: Once again, we prove only that $[\alpha\cdot\alpha^{\leftarrow}]=[c_x]$. Indeed, define $H:[0,1]^2\to X$ by

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$H(s,t)=\begin{cases}x & \mbox{if}\quad 0\leqslant s\leqslant\frac{t}{2}\\ \alpha(2s-t) & \mbox{if}\quad \frac{t}{2}\leqslant s\leqslant\frac{1}{2}\\ \alpha(2-2s-t) & \mbox{if}\quad \frac{1}{2}\leqslant s\leqslant 1-\frac{t}{2}\\ x & \mbox{if}\quad 1-\frac{t}{2}\leqslant s\leqslant 1\end{cases}$

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Once again, one can verify that all the boundary conditions are in agreement, and so $H$ is continuous by the gluing lemma. Moreover, one can verify that, indeed, $H(s,0)=\alpha\cdot\alpha^{\leftarrow}$ and that $H(s,1)=c_x$. Thus, $\alpha\cdot\alpha^{\leftarrow}\simeq c_x$ and so $[\alpha]\cdot[\alpha^{\leftarrow}]=[\alpha\cdot\alpha^{\leftarrow}=[c_x]$.

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Last, but certainly not the least annoying, is the verification that (like all good products should be) the multiplication is associative. Indeed:

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Theorem: Let $x,y,z,w\in X$ and $\alpha\in\mathcal{P}(x,y),\beta\in\mathcal{P}(y,z),\gamma\in\mathcal{P}(z,w)$. Then,

$([\alpha]\cdot[\beta])\cdot[\gamma]=[\alpha]\cdot([\beta]\cdot[\gamma])$

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Proof: As per usual, we define a homotopy $(\alpha\cdot\beta)\cdot\gamma\simeq\alpha\cdot(\beta\cdot\gamma)$. In particular, define $H:[0,1]^2\to X$ according to the rule

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$H(s,t)=\begin{cases}\alpha\left(\frac{4s}{t+1}\right) & \mbox{if}\quad 0\leqslant s\leqslant\frac{t+1}{4}\\ \beta(4s-t-1) & \mbox{if}\quad \frac{t+1}{2}\leqslant s\leqslant\frac{t+2}{4}\\ \gamma\left(\frac{4s-t-2}{2-t}\right) & \mbox{if}\quad \frac{t+2}{4}\leqslant s\leqslant 1\end{cases}$

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I, once again, leave the busy work to the reader to check that the conditions agree on overlap, and thus $H$ is continuous. Moreover, I leave it to you lucky folks to check that, in fact, $H(s,0)=(\alpha\cdot\beta)\cdot\gamma$ and $H(s,1)=\alpha\cdot(\beta\cdot\gamma)$. $\blacksquare$

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Now that we have verified that our product satisfies such nice properties, we can, in fact, create a fairly tame structure. In particular, call a category a groupoid if every arrow is an isomorphism. We claim then that if $\Pi_1(X)$, called the fundamental groupoid of $X$, is the category whose object set $X$, for which $\text{Hom}_{\Pi_1(X)}(x,y)=\mathcal{P}_X[x,y]$, and for which composition is taken by the product $\cdot$ then:

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Theorem: $\Pi_1(X)$ is a groupoid.

Proof: It’s evident that $\Pi_1(X)$ is a category, whose identity at each object $x$ is merely $[c_x]$. We see then that each arrow $[\alpha]\in\mathcal{P}[x,y]$ is indeed invertible since $[\alpha^{\leftarrow}]\in\mathcal{P}[y,x]$ is an inverse. $\blacksquare$

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References:

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[1] Spanier, Edwin Henry. Algebraic Topology. New York: McGraw-Hill, 1966. Print.

[2] Hatcher, Allen. Algebraic Topology. Cambridge: Cambridge UP, 2002. Print.

[3] Bredon, Glen E. Topology and Geometry. New York: Springer-Verlag, 1993. Print.