# Is every isometric immersion between surfaces of equal area injective?

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.

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.