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.

Answer 2

We can generalize to get the cases not covered by Eric Wofsey's response. We can prove that if $X$ is path-connected, then its fundamental group is trivial.

We define $\kappa := |X|$.

Lemma. The three following properties are equivalent:

1. We can cover $[0,1]$ with $\kappa$-many disjoint closed non-full sets
2. We can cover $]0,1[$ with $\kappa$-many disjoint sets which are closed in $[0,1]$
3. $X$ is path-connected.

Proof. We already know that 1 and 3 are equivalent. It is clear that 2 => 1 since you can just add a point at each extremity, so we will prove that 1 => 2. Let $(C_i)_{i \in \kappa}$ be a collection of closed sets that witnesses this property. Without loss of generality, we can suppose that $0,1 \in C_0$. Then there exists some $a,b \in C_0$ such that $a < b$ and $]a,b[ \ \cap \ C_0 = \emptyset$. The collection $(C_i\ \cap\ ]a,b[)$ is a collection of sets closed in $[a,b]$ which cover $]a,b[$, and therefore we have property 2. QED.

Now, let's suppose that $X$ is path-connected, and let $f : [0,1] \to X$ be a loop at some point $x_0$. We need to find a continuous map $F : [0,1]^2 \to X$ such that

1. $F(t,0) = f(t)$
2. $F(0,t) = F(1,t) = x_0$
3. $F(t,1) = x_0$

To construct this function, since the values are already determined on the sides of the square, we need to cover $]0,1[^2$ with $\kappa$-many disjoint sets which are closed in $[0,1]^2$. Using the lemma, we get a collection $(C_i)$ covering $]0,1[$. Then the collection $(C_i \times C_j)_{i,j}$ covers $]0,1[^2$, has size $\kappa^2 = \kappa$, and all these sets are disjoint and closed in $[0,1]^2$.