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The groupoid version of van Kampen's theorem states that if $X$ is a topological space and $X = U\cup V$ where $U,V$ are open, then the fundamental groupoid $\tau_{\leq 1}(X)$ of $X$ is the limit of the diagram $$\tau_{\leq 1}(U)\leftarrow \tau_{\leq 1}(U\cap V)\rightarrow \tau_{\leq 1}(V)$$ where the morphisms between fundamental groupoids above are those induced by the inclusions of $U\cap V$ into $U$ and $V$.

If $S = S^1$, we can choose $U$ and $V$ to be homeomorphic to copies of $(0,1)$, so that $U\cap V = (0,1)\sqcup (0,1)$. We can't use the usual version of van Kampen's theorem here because $U\cap V$ isn't path connected. But is it doable to show that $\pi_1(S^1)=\mathbb Z$ using the groupoid version? If so, can I see a proof? Thanks.

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One really needs to set up the Seifert-van Kampen theorem for the fundamental groupoid $\pi_1(X,S)$ on a set of base points chosen according to the geometry. One sees the circle as obtained from the unit interval $[0,1]$ by identifying $0$ and $1$. So we need the fundamental groupoid on the set $S= \{0,1\}$. Thus $\pi_1(X)$ can be too large; $\pi_1(X,x)$ can be too small; but $\pi_1(X,S)$ can be just right.

For example we may have $X= U \cup V$, with $X,U,V$ path connected space but $U \cap V$ may have $1,024$ path components. It is also often the case that choosing a single base point will neglect any symmetry in the geometric situation.

Full details are in the book Topology and Groupoids. The first step is to prove a pushout form of the SvKT, when $S$ meets each path component of $U,V, W =U \cap V$. Assume further that $V$ say is path connected, and $x \in W \cap S$. Then one can set up a composite of pushouts:

$$\begin{matrix}\pi_1(W,S) & \to & \pi_1(V,S) & \to & \pi_1(W,x)\\ \downarrow & & \downarrow & &\downarrow\\ \pi_1(U,S) & \to & \pi_1(X,S)& \to & \pi_1(X,x) \end{matrix} $$ which is itself a pushout.

In the case of the circle this becomes a composite of the form

$$\begin{matrix} \{0,1\} &\to & \{ 0 \}\\ \downarrow & & \downarrow\\ \pi_1([0,1], \{0,1\}) & \to & \pi_1(S^1,0). \end{matrix} $$

Since $[0,1]$ is convex, $\pi_1([0,1], \{0,1\}) \cong \mathcal I$, where the latter is the groupoid with two objects $0,1$ and two non identity arrows $\iota:0 \to 1$ and its inverse $ \iota^{-1}:1 \to 0$. This is a model of the "unit interval" in the category of groupoids. From these facts we can determine $\pi_1(S^1,0)$. Note that within $[0,1]$ we have no special reason to choose $0$ or $1$ as base point, so we choose them both.

Categorically speaking, the groupoid $\mathcal I$ has similar properties in the category of groupoids to those of the integers $\mathbb Z$ in the category of groups. So although it looks "trivial", it should not be ignored.

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