Does $S^1 \vee S^1$ have a cover homotopically equivalent to the wedge sum of $k$ circles?

Question

I'm stuck on the following problem: Is it true that for every $k$ the space $S^1 \vee S^1$ has a covering space homotopically equivalent to the wedge sum of $k$ circles? I don't really have a good intuition on covering spaces or the difference between homotopy and homemorphisms so any tips on approaching this would be appreciated.

Answer 1

For $k\geq 2$ here is a way to get them. One class of covering spaces for $S^1\vee S^1$ comes from covering spaces of one of the $S^1$'s and attaching copies of the other $S^1$ at each lifted base point. That is, a space which is an $S^1$ along with $n\geq 1$ copies of $S^1$ wedged to it at different points. This space is homotopy equivalent to $n+1$ wedged $S^1$'s by way of collapsing $1$-cells.

For $k=0$, the universal cover of $S^1\vee S^1$ is contractible, so it is homotopy equivalent to the wedge sum of zero $S^1$'s.

For $k=1$, you can take the covering space associated with the subgroup generated by one of the generators. This looks like a circle with two "leaves" of the Cayley graph for $F_2$ attached. Each leaf is contractible to the basepoint.