TITLE: Does nonexpanding map between manifolds decrease volume?
QUESTION [7 upvotes]: (This question is a special case of a question I asked at SE, which got no answer there) 
Let $M,N$ be diffeomorphic connected compact Riemannian manifolds, and let $f:M \to N$ be a surjective nonexpanding map (i.e Lipschitz with constant $1$). 
Assume that $f$ is strictly nonexpanding, i.e there exists $p,q \in M$ such that $$d(f(p),f(q)) < d(p,q)$$
Is it true that $\operatorname{Vol}(N)<\operatorname{Vol}(M)$?
Edit: 
I want $M,N$ to be smooth manifolds with or without boundary. 
Nik's answer gives a counter example (when $\operatorname{Vol}(N)=\operatorname{Vol}(M)$), but I am not sure it is smooth in the sense mentioned above. (The same problem comes up in this example given at S.E).

Note that if we do not assume $M,N$ are diffeomorphic then the answer is negative:
A counter-example is $f:[0,2\pi] \to \mathbb{S}^1, f(t)=e^{it}$. 
Partial result:
In the case the manifolds are one-dimensional the answer is positive:
Assume otherwise. Then $\operatorname{Vol}(N)=\operatorname{Vol}(M)$. Since every two compact connected one-dimensional Riemannian manifolds with equal volumes are isometric, there exists an isometry $\phi:N \to M$.
Thus, $f \circ \phi:N \to N$ is surjective and nonexpanding, hence an isometry.

REPLY [2 votes]: I don't think so, if "compact manifold" means "compact manifold with boundary" (as it seems to from your example) and "diffeomorphic" means "with diffeomorphic interiors". My previous example wasn't right, as Willie Wong pointed out, because after gluing the result wasn't Riemannian. But we can fix that problem by a simple modification: let $M$ be the unit disk minus the region $y > \sqrt{|x|}$, and let $f$ be the map that zips up the slit. That works, doesn't it?