Abstract Nonsense

Crushing one theorem at a time

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.

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August 30, 2012 - Posted by | Algebraic Topology, Topology | , , , , , , , ,

1 Comment »

  1. […] The Fundamental Groupoid and Group (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by The Fundamental Groupoid and Group (Pt. III) « Abstract Nonsense | August 30, 2012 | Reply


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