Can one reconstruct a topological space by knowing all of its covering spaces? Let $X$ be a topological space with a universal cover $\tilde{X}$. Let $X_{i}$ be the covering spaces of $X$ excluding $X$ and let $p_{ij}: X_{i}\to X_{j}$ be covering maps if such a map exists for $i,j$. Is $X$ the colimit of this diagram? 
EDIT: One needs to add a condition to $\pi_{1}\left(X\right)$. As a group it needs to be generated by its proper subgroups. For instance the projective plane has only one covering space which is the universal cover. 
 A: I believe the answer is yes, once the question is modified as follows: work in pointed spaces, take the $p_{i,j}$ to be covering maps over $X$, and take only connected coverings. The first and last assumptions are not strictly necessary, see the end.
Consider the diagram described in the question: it consists of all connected covering spaces and all covering maps over $X$. Now take the diagram with the same objects, but only with maps from $\tilde{X}$. Note that all maps in the diagram are quotient maps. The colimit of such a diagram in topological spaces is a quotient, since a map from the diagram to a space $Y$ is precisely the same as one from $\tilde{X}$ which is well-defined on each quotient space, i.e. which is constant on each fiber of any one of the maps ("being in the same fiber of a given map" is an equivalence relation; taken together, the maps generate one equivalence relation).
That the colimit is $X$ follows by noting that the equivalence relation by which we quotient $\tilde{X}$ contains all pairs $(p, g.p)$ for $g \in \pi_1(X)$ and $p\in\tilde{X}$.
Now consider the complete diagram, with all maps. The colimit is the same as the previous one considered: to see this, check that any map from the sub-diagram described above extends uniquely to the full diagram.
PS: Regarding the assumptions near the beginning - without the first assumption, one has automorphisms (deck transformations) which should make the colimit with just $\tilde{X}$ already equal $X$. Without the second, you get a wild thing (circles cover each other in all kinds of funny ways, so $S^1$ already gives strange behavior...) Without the third, well, it still works, but I don't think it's as nice. The proof is easy though: the group assumption becomes unnecessary because $X\sqcup X$ is a covering space of $X$ and has two nice maps from the universal cover.
