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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy