I have a question on the standard algebraic topology proof that a subgroup of a free group is free. My understanding of that proof (mostly from Hatcher's topology) is as follows:
- We define a topology on a graph as a $1$-complex.
- We show that the fundamental group of a graph is the free group generated by one symbol for each loop in the graph.
- We show that a covering space of a graph is again a graph.
- We then argue as follows: Given a free group $F$, construct a graph $G$ whose fundamental group is isomorphic to $F$ (as Hatcher notes, this graph can be a wedge of circles). Then using the theory of covering spaces, any subgroup $H \leq F$ corresponds to the fundamental group of some covering space of $G$, which is a graph by item 3. Thus $H$ is a free group by item 2.
My concern is whether this applies when $F$ is infinitely generated - in order to apply the standard results from covering spaces, don't we need to know that the graph $G$ we construct is semilocally simply connected? But (at least visually) if $F$ is a countably infinite set, I image that $G$ looks like the infinite earring, which is famously not semilocally simply connected.
I suspect that there is some difference I am missing between the topology of the infinite earring as a subset of $\mathbb{R}^2$ and the topology of a graph on one vertex with infinitely many loops at that vertex as a $1$-complex, but I'm not sure I know enough about CW-complexes to see what that difference is.
