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There is a nice result that says for any group $G$, there exists a topological space $X$ such that the fundamental group $\pi_{1}(X)$ of $X$ is isomorphic to $G$. The proof is not terribly complex: we construct a space whose fundamental group is a suitable free group $F(S)$ by gluing circles (one for each generator of $F(S)$) to a common base point, and then we quotient out by relations by attaching 2-discs to the corresponding circles to obtain a space with fundamental group $G$.

I was recently told that the corresponding result does not hold for groupoids and fundamental groupoids; that is, there exists some groupoid $\mathcal{G}$ which is not isomorphic to the fundamental groupoid $\pi_{\leq 1}(X)$ of any space $X$.

My question is, is this true? If so, what is an example of a groupoid that is not a fundamental groupoid?

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The group $\mathbb{Z}$, interpreted as a one-object groupoid, is not the fundamental groupoid of any topological space $X$. The number of objects of a fundamental groupoid is the number of points of the topological space, which would have to be one in this case. But there's only one one-point topological space and its fundamental groupoid corresponds to the trivial group.

The right generalization of the fundamental group result in this context in that each groupoid is isomorphic to the fundamental groupoid of a topological space $X$ relative to a certain choice of basepoints. To see this, we may decompose the groupoid into its connected components. Then, each component is determined up to isomorphism by its number of elements and automorphism group. Choose a connected topological space whose fundamental group is this automorphism group and the fundamental groupoid of this space with respect to any set of basepoints that has the right cardinality* will realize the corresponding connected component of the groupoid we started with.

*There's a subtle point to be made here about enough points existing, but we can choose our space to have arbitrarily large cardinality without affecting the fundamental group by taking the product with enough copies of $\mathbb{R}$ - of course, we do have to require our groupoid to be small in the sense of category theory.

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  • $\begingroup$ Thank you! That is quite a nice result as well. $\endgroup$
    – ckefa
    Jan 21 at 21:22

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