Can a space with $\pi_1(X)=F_2$ have a nontrivial covering space which is homeomorphic to $X$? Any finite sheeted connected cover of the circle is again homeomorphic to a circle. On the group level, this is consistent with the fact that subgroups of $\mathbb Z$ are isomorphic to $\mathbb Z$. This leads to the question of whether one could find nontrivial covers which are still homeomorphic to the base space for other groups that have subgroups isomorphic to themselves. In particular, consider the free group on 2 generators $F_2$. It has plenty of subgroups isomorphic to $F_2$, so one can ask
Question Is there a space $X$ with $\pi_1(X)\cong F_2$ which has a nontrivial cover $p\colon E\to X$, where $E\cong X$?
Clearly the standard example of $X=S^1\vee S^1$ doesn't work, but perhaps starting with $X$ as a graph of infinite valence might work, or perhaps there is an even more clever construction.
 A: Let $G\subset F_2=\langle a,b\rangle$ be the subgroup generated by $a^2$ and $b^2$.  Let $X_0=S^1\vee S^1$ and recursively define $X_{n+1}$ to be the covering space of $X_n$ associated to $G$, identifying $\pi_1(X_n)\cong G$ with $F_2$.  Note that there is an embedding $X_0\to X_1$ that induces an isomorphism on $\pi_1$, mapping the two loops of $X_0$ to the loops in $X_1$ corresponding to $a^2$ and $b^2$.  Taking the covering spaces associated to $G$ on both sides, this induces an embedding $X_1\to X_2$, and then $X_2\to X_3$, and so on.  Let $X$ be the colimit of this sequence of embeddings $X_0\to X_1\to X_2\to\dots$.
Since each of the maps $X_n\to X_{n+1}$ is a closed embedding, $\pi_1(X)$ is the colimit of the groups $\pi_1(X_n)$ and thus is isomorphic to $F_2$.  Now, the covering maps $p_n:X_{n+1}\to X_n$ are compatible with the inclusion maps in the colimit, and thus induce a map $p:X\to X$ which induces the inclusion $F_2\cong G\to F_2$ on $\pi_1$.  So, if this colimit map $p$ is still a covering map, it is the desired example.
To sketch a proof that $p$ is a covering map, fix a pair of points $x,y\in X_0$, one in each circle, and neither one a dyadic rational fraction of the way around the circle.  This means that neither $x$ nor $y$ will ever become a vertex in the natural CW-complex structure on any $X_n$.  Then for all $n$, $X_n\setminus\{x,y\}$ is evenly covered by $p_n$ (identifying $x$ and $y$ as points of $X_n$ via the embedding $X_0\to X_n$).  It follows then that $X\setminus\{x,y\}$ is evenly covered by $p$.  Varying $x$ and $y$, sets of the form $X\setminus\{x,y\}$ cover $X$.
(Intuitively, $X$ looks like $S^1\vee S^1$ with infinitely dense tree-like "fuzz" attached at each dyadic rational along each circle, where the fuzz itself has fuzz attached at each dyadic rational along each edge, and then the fuzz's fuzz also has fuzz, and so on.)
A: So I think the answer is simple. Start with $S^1\vee S^1$ and take any cover for which $\pi_1$ of that cover is a free group of rank 2. Now thicken these examples slightly by replacing each vertex by an open disk and each edge by an open strip. So $X$ is homeomorphic to an open disk with two closed disks removed. The cover of $X$ will also include thickenings of the tree parts of the graph, but because we are not including the boundary, these will just be homeomorphic to open disks glued to the rest of the graph.
