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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

dmontana
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    Well, assuming if your space has cardinality $\geq \mathfrak{c}$ you can do the same argument and see that any injective map $f: I^2\to X$ will be continuous, so the fundamental group is trivial. – Maxime Ramzi Feb 03 '19 at 13:55
  • What happens if we already have a bijection $I \rightarrow X$? Can we construct an injection $I^2 \rightarrow X$? – dmontana Feb 03 '19 at 14:16
  • NVM, it isn't a loop then – dmontana Feb 03 '19 at 14:23
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    I might have been a bit quick indeed, but note that "injectivity" is a much stronger condition than continuity; it suffices that every point has a finite number of antecedents, and this will get you closer to an actual solution – Maxime Ramzi Feb 03 '19 at 14:32

1 Answers1

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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!).

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.

Eric Wofsey
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