Fundamental group of uncountable space with cofinite topology

Question

So it is easy to show that such a space is path connected (assuming CH, injective maps $f:I \rightarrow X$ are continuous) but I'm not sure how to start computing the fundamental group. Will it depend on the cardinality?

Thank you

Answer 1

Assuming $|X|\geq 2^{\aleph_0}$, then $X$ is contractible so its fundamental group is trivial. To prove this, let $f:X\times(0,1)\to X$ be an injection (here we use the fact that $|X|\geq 2^{\aleph_0}$). Define $H:X\times[0,1]\to X$ by $H(x,0)=x$, $H(x,1)=x_0$ for some fixed point $x_0\in X$, and $H(x,t)=f(x,t)$ if $t\in (0,1)$.

I claim $H$ is continuous, and thus a contraction of $X$. To show this, it suffices to show $H^{-1}(\{x\})$ is closed for each $x\in X$. If $x\neq x_0$, then $H^{-1}(\{x\})=\{(x,0)\}\cup f^{-1}(\{x\})$ which is a finite set and hence closed. If $x=x_0$, then we have $H^{-1}(\{x_\})=\{(x_0,0)\}\cup f^{-1}(\{x_0\})\cup X\times\{1\}$ which is a union of finite sets and the closed set $X\times\{1\}$ and thus also closed. Thus $H$ is continuous, and $X$ is contractible.

See What is the homotopy type of the affine space in the Zariski topology..? for an interesting generalization of this argument to some related spaces that arise naturally in algebraic geometry (though their fundmental groups do not arise naturally!).

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If you don't assume $|X|\geq 2^{\aleph_0}$ (or CH), then I don't know what can be said in general. It is consistent for $X$ to be totally path-disconnected if $|X|<2^{\aleph_0}$ (see https://mathoverflow.net/a/48991/75) and so the fundamental group with any basepoint will still be trivial. But if $|X|<2^{\aleph_0}$ and $X$ is not totally path-disconnected, I don't know what can be said about its fundamental group.