(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.