TITLE: Self-covering spaces
QUESTION [15 upvotes]: Let $M$ be a connected Hausdorff second countable topological space. I will call $M$ self-covering if it is its own $n$-fold cover for some $n>1$. For instance, the circle is its own double cover by $e^{2\imath\varphi}\mapsto e^{\imath\varphi}$. 
Questions:

What would be a reasonable assumption on $M$ to ensure it is self-covering? Is every connected compact/closed topological manifold self-covering?
Assume that $M$ is a differentiable manifold, and let $\phi:M\to M$ be the local diffeomorphism providing the $n$-fold cover for $n>1$. Is it true that the modulus of the Jacobian $|\det J(d\phi(x))|<1$ everywhere aside from countably many isolated points $x$ where it can be 1? In other words, is a covering map a volume contraction?

Please, note that I am not a topologist; more accessible language and references would be appreciated.

REPLY [17 votes]: Here are some pretty examples of self covering manifolds: Suppose that $F$ is a manifold and $f \colon F \to F$ is a periodic homeomorphism of period $k$.  We define $M_f = F \times I \,/\, f$ to be the mapping torus with monodromy $f$.  That is, form the product $F \times I$ and identify the two ends via $f$.  Note that $M_f$ is an $F$-bundle over the circle. 
Now, the $(k+1)$-fold cover of $M_f$, obtained by "unwrapping the circle direction" is the mapping torus for $f^{k+1} = f$.  Thus the $(k+1)$-fold cover is homeomorphic to $M_f$.  
As a concrete example, the trefoil knot complement $X_T$ is a once-punctured torus bundle over the circle, with monodromy of order six.  Thus $X_T$ seven-fold covers itself.  (And also five-fold covers itself.)  This trick works for any torus knot complement.