Maps between manifolds with boundary and homeomorphism

Question

Assume we have $f:(M,\partial M)\rightarrow (N,\partial N)$ connected 3-manifolds, not compact, such that $f$ is an homeomorphism onto its image and $f(\partial M)=\partial N$. Can say that $f$ has to be surjective?

Answer 1

No, let the boundaries be empty, and assume that $M$ is a ball. Then this does not hold. If you want boundaries, you can make multiple components, but one component (of both $M$ and $N$) without a boundary and repeat this trick.

If you are not satisfied by this, you can take $N$ with a non-trivial boundary and remove an closed ball $B$. Then $M=N-B$ is a manifold that is clearly embedded in $N$ and is surjective on the boundary.

Answer 2

Take $M=N=[0,\infty)$ and take any continuous injection $$f : M=[0,\infty) \to [0,1) \subset [0,\infty) = N $$ such that $f(0)=0$.