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Let $M,N$ be smooth connected, compact two-dimensional Riemannian manifolds, such that $M$ has a non-empty Lipschitz boundary. Suppose that $\operatorname{Vol}(M)=\operatorname{Vol}(N)$.

Question: Let $f:M \to N$ be a smooth isometric immersion (i.e $df_p$ is an isometry for every $p \in M$). Must $f$ be surjective?

This equivalent to $f$ being injective "a.e. in the image"- i.e. $|f^{-1}(q)| \le 1$ for a.e. $q \in N$. (see below).

The argument given here shows that if $\partial M=\emptyset$, then $f$ is surjective.

Proof of the equivalence:

By the area formula $$ \text{Vol}(M) = \int_M 1=\int_M \det df = \int_N |f^{-1}(y)|=\int_{f(M)} |f^{-1}(y)|. $$ So, if $|f^{-1}(y)| \le 1$ a.e. on $N$, then $ \text{Vol}(N)=\text{Vol}(M) = \text{Vol}(f(M))$. On the other hand, if $\text{Vol}(f(M))=\text{Vol}(M)$, then $$\text{Vol}(f(M))=\text{Vol}(M)= \int_{f(M)} |f^{-1}(y)| \ge \int_{f(M)} 1= \text{Vol}(f(M)), $$ so $|f^{-1}(y)| \le 1$ a.e. on $f(M)$, hence also on $N$.

We proved that $|f^{-1}(y)| \le 1$ a.e. on $N$ if and only if $\text{Vol}(f(M))=\text{Vol}(N)$.

Since $f(M) $ is compact, being of full measure in $N$ is equivalent to being equal to $N$.

Comment:

Some amount of non-injectivity is clearly possible:

Take for example $M=[-1,1]^2$, and let $N=M/\sim$ be the flat $2$-torus with $\sim$ the standard equivalence relation. Then the quotient map $\pi:M\to N$ is not everywhere injective.

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Let $M= \mathbb D = \{(x, y) \in \mathbb R^2 : x^2 + y^2 \le 1\}$ and $N = [-\pi, \pi] \times \mathbb R /\sim$, where $\sim $ identify $(x, y)$ with $(x, y+ n/2)$ for all $n\in \mathbb Z$. Give $M, N$ the standard Euclidean metrics in $\mathbb R^2$. Then $M, N$ has the same volume. Let $f$ be the composition

$$ M \overset{i}{\to} [-\pi, \pi] \times \mathbb R \overset{\pi}{\to} N.$$

Then $f$ is an isometric immersion which is not injective and not surjective.

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    $\begingroup$ Thanks! I think I get your idea- you are rolling a disk over a cylinder, right? Very cool! $\endgroup$ Aug 25, 2020 at 14:04
  • $\begingroup$ Maybe you will have an idea about this follow-up question-math.stackexchange.com/questions/3802867/…. Thanks again for your great example. $\endgroup$ Aug 25, 2020 at 14:26

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