# Fundamental Group of a finite set with discrete topology

let S be a finite set with say n elements. Give it the discrete topology, Now what can we say about its fundamental group? Atleast can we determine the fundamental group of a set with two elements? Thanks

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Such a space is not path-connected, so you have to keep track of the basepoint. Each path-connected component is trivially contractible (it consists only of one point). The fundamental group is trivial for each component. You don't need the finiteness of $S$.

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Yes, I missed out the fact that image of a connected domain is connected for a continuous map. Hence every loop in S should be a constant map. From which it turns out that the fundamental group is trivial. Thanks! – Dinesh Apr 23 '11 at 17:52

You need to choose a base point both in $S^1$ and $S$. Since $S^1$ is connected, it must be mapped to this base point entirely. Hence there is only one base-point preserving loop in $S$, and thus the fundamental group is trivial. This applies to any set with the discrete topology, no matter if it is finite or not.

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@Theo: what's $S^1$? The unit circle? – Rudy the Reindeer Apr 25 '11 at 8:57
@Matt: Yes. ${}$ – t.b. Apr 25 '11 at 8:58
@Theo: I assume you are using $S^1$ instead of $[0,1]$ because the question is about closed paths. Why do you have to choose a base point in the domain of the paths? Shouldn't you say "...since $S^1$ is connected, it must be mapped to this base point entirely....", where that base point is the one in $S$? – Rudy the Reindeer Apr 25 '11 at 9:09
@Matt: I'm just using $S^1 \cong [0,1]/\{0 \sim 1\}$. The chosen base point $\ast$ on $S^1$ corresponds to the image of the identified end points of the interval. In other words, there is a bijection between loops $\gamma: [0,1] \to S$ (i.e. maps s.t. $\gamma(0) = \gamma(1) = s_0$) and base-point preserving maps $\gamma: (S^1, \ast) \to (S,s_0)$. – t.b. Apr 25 '11 at 9:14
@Theo: thank you, now I understand. I didn't understand why it was necessary to pick base points when one can show that every point in $S$ can only have the constant loop. It's necessary because that is how the fundamental group is defined. – Rudy the Reindeer Apr 25 '11 at 10:43