2

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

Dinesh
  • 1,737

2 Answers2

9

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 $S$.

Thomas Rot
  • 10,023
  • 1
    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
  • To calculate the fundamental group of $\pi_1(\mathbb Z,n_0)$ where $n_0\in$ for each $n_0\in \mathbb Z$, we can say that it is the trivial group $ [e]$ according to your argumentation, am I correct? – Verónica Rmz. Sep 17 '19 at 16:37
4

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.

t.b.
  • 78,116
  • @Theo: what's $S^1$? The unit circle? – Rudy the Reindeer Apr 25 '11 at 08:57
  • @Matt: Yes. ${}$ – t.b. Apr 25 '11 at 08:58
  • @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 09:09
  • 1
    @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 09:14
  • @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
  • @Matt: yes, exactly. Maybe I should have pointed that out... – t.b. Apr 25 '11 at 10:46
  • So with specified base point in $S^1$ can we define the fundamental group as the maps from $S^1$ to $S$ modulo homotopy? I mean to ask can we completely replace $[0,1]$ with $S^1$ in an equivalent definition of fundamental group? – Dinesh Apr 29 '11 at 01:36
  • @Dinesh: Yes, provided you replace "homotopy" by "pointed homotopy". – t.b. Apr 29 '11 at 01:44