Question: Is there a second-countable, connected, locally path-connected, semi-locally simply connected (and "perhaps" Hausdorff) topological space $X$ such that $\#\pi_1(X)>\aleph_0$?

Uninteresting story of the question: I was doing an exercise which asked to prove that a Hausdorff, locally compact, second-countable, connected topological manifold must have countable fundamental group. My idea was observing that, by Poincaré-Volterra theorem, its universal cover is second-countable, so its fibres are countable.

However, the exercise hinted towards a more direct, and somehow visual, proof, which I decided to follow. Jokes on me, the work I did implies (in my opinion) way too much. In fact, just by "following the instructions", I ended up using only N2 and the "triple connection" hypothesis that guarantees the existence of the universal covering. Not even T2, actually. I wanted to be sure I was wrong, before diving back in, though.


The fundamental group must be countable in your situation. Here's a proof, which has a good visualization in my head from which I concocted it, although in the end it's easier to write out the proof than to draw the picture.

From your hypotheses, $X$ has a countable basis $\{U_i\}_{i=1}^\infty$ consisting of path connected open sets for which the inclusion induced homomorphism $\pi_1(U_i) \to \pi_1(X)$ is trivial. Also, given two basis elements $U_i,U_j$, their intersection consists of countably many path components, denote them $$U_i \cap U_j = \cup_{k=1}^\infty V_{ij}^k $$ Pick points $p_i \in U_i$ and $q_{ij}^k \in V_{ij}^k$. Declare one of the $p$'s to be the base point, say $p_1 \in U_1$.

For any closed path $\gamma : [0,1] \to X$ based at $p_1$, by the Lebesgue number lemma we may subdivide $$0=x_0 < x_1 < ... < x_M=1 $$ so that for each $m=1,...,M$ the path $\gamma[x_{m-1},x_m]$ has image in one of the $U_i$'s, call it $U_{i_m}$. We'll assume $i_1=i_M=1$.

Let's do a preliminary path homotopy on $\gamma$, which will achieve the following effect: denoting $y_m = \frac{x_{m-1}+x_m}{2}$ which is the midpoint of the interval $[x_{m-1},x_m]$, we may assume that $\gamma(y_m) = p_{i_m}$ for $2 \le m \le M-1$. To achieve this, cut $\gamma$ at $y_m$ and then insert a new path which first travels along some path in $U_{i_m}$ from $\gamma(y_m)$ to $p_{i_m}$ and then backwards along the same path.

Next, note that we have a nonempty intersection $U_{i_{m-1}} \cap U_{i_m} \ne \emptyset$ because that set contains the point $\gamma(x_m)$. Let $V_{i_{m-1} i_m}^{k_m}$ be the path component of that intersection that contains $\gamma(x_m)$.

Now I'll construct a countable collection of closed "model paths" based at $p_1$, and I'll pick out one of those paths which is path homotopic to $\gamma$.

For each $i,j,k$ such that $V_{ij}^k \ne \emptyset$ let $\delta_{ij}^k$ be the concatenation of a path in $U_i$ from $p_i$ to $q_{ij}^k$ with a path in $U_j$ from $q_{ij}^k$ to $p_j$. Since the inclusions from $U_i$ and $U_j$ into $X$ induce trivial maps on fundamental groups, it follows that the path homotopy class of $\delta_{ij}^k$ is well-defined. There are countably many of the $\delta$'s, and so the collection of paths obtained by concatenating a finite sequence of the $\delta$'s is countable. These are the "model paths".

So now we just have to show that $\gamma$ is path homotopic to the path $$\delta_{i_1i_2}^{k_2} * ... * \delta_{i_{m-1}i_m}^{k_m} $$ For that purpose, for each $m=1,...,M-1$ we pick a path $\eta_m$ in $V_{i_{m-1}i_m}^{k_m}$ from the point $\gamma(x_m)$ to the point $q_{i_{m-1}i_m}^{k_m}$, and then we cut $\gamma$ at $x_m$ and insert a copy of $\eta_m \bar\eta_m$. It follows that $\gamma$ is path homotopic to $$\underbrace{(\gamma[x_0,x_1] \, \eta_1 \, \bar\eta_1 \, \gamma[x_1,y_2])}_{\delta_{i_1i_2}^{k_2}} * \underbrace{\gamma[y_2,x_2] \, \eta_2 \, \bar\eta_2 \, \gamma[x_2, y_3])}_{\delta_{i_2i_3}^{k_3}} * ... * \underbrace{(\gamma[y_{m-1},x_{m-1}] \, \eta_{m-1} \, \bar\eta_{m-1}\, \gamma[x_{m-1,}x_m])}_{\delta_{i_{m-1}i_m}^{k_m}} $$

  • $\begingroup$ Ok, it's the same idea of the hint and almost verbatim what I had done. Thanks. My doubts were on the fact that I did not understand why the author bothered doing it in that special case only (since he had summarized the essential facts of cover theory beforehand). $\endgroup$ – user228113 Aug 6 '17 at 17:37

This is a job for topological fundamental groups. In fact one can weaken the hypotheses a little bit to include some non-locally path-connected spaces.

Theorem: If $X$ is second countable and admits a simply connected covering space, then $\pi_1(X,x)$ is countable.

Let $\Omega(X,x)$ be the based loop space of $X$ with the compact-open topology. Let $\pi_{1}^{qtop}(X,x)$ be the fundamental group equipped with the quotient topology with respect to the map $q:\Omega(X,x)\to\pi_{1}^{qtop}(X,x)$, $q(\alpha)=[\alpha]$ identifying homotopy classes of loops. Warning: $\pi_{1}^{qtop}(X,x)$ is a quasitopological group (in particular is homogeneous) but may not be a topological group.

Claim 1: $\pi_{1}^{qtop}(X,x)$ is separable.

Proof. Since $S^1$ is locally compact Hausdorff and $X$ is second countable, $\Omega(X,x)$ is second countable (see Engelking 3.4.16). Every second countable space is separable and every continuous image of a separable space is separable. Therefore $\pi_{1}^{qtop}(X,x)$ is separable.

Claim 2: $\pi_{1}^{qtop}(X,x)$ is discrete.

Proof. The details are a bit technical but it is known that a covering map (and even more general a semicovering map) $p:\tilde{X}\to X$ induces a homomorphism $p_{\#}:\pi_{1}^{qtop}(\tilde{X},\tilde{x})\to \pi_{1}^{qtop}(X,x)$ which is also an open embedding. If $p$ is the universal cover, then $\pi_{1}^{qtop}(\tilde{X},\tilde{x})=1$ is the trivial discrete group. Since the trivial subgroup $p_{\#}(\pi_{1}^{qtop}(\tilde{X},\tilde{x}))=1$ is open and $\pi_{1}^{qtop}(X,x)$ is homogeneous, $\pi_{1}^{qtop}(X,x)$ is a discrete group. Under your original assumptions the simpler arguments in this paper suffice to prove Claim 2.

Proof of Theorem. Every separable discrete space is countable.


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