If $f : X \to Y$ is a continuous map between manifolds with boundary and $f(\partial X) \subseteq \partial Y$, is $f$ surjective?

Question

Let $X$ and $Y$ be $n$-dimensional manifolds with boundary and $f:X \to Y$ be a continuous function.

Suppose that $f(\partial X) \subseteq \partial Y$. Does this imply surjectivity and that $f(\partial X) = \partial Y$?

Follow up question: Do the implications hold when $f$ is a local homeomorphism (or weaker still an open map?)

Answer 1

Not even when $f$ is a local homeomorphism: consider the inclusion $[0,1] \times \{0\} \subset [0,1] \times \{0,1\}$. For a connected (but non-compact) example, consider $(0,1) \subset [0,1]$.

However when $X$ is compact and $Y$ is connected (and both are Hausdorff), then a local homeomorphism $X \to Y$ is always surjective, and this isn't restricted to manifolds (I might as well include the proof: $f$ is an open map hence $f(X)$ is open, but $X$ is compact so $f(X)$ is compact, hence closed in the Hausdorff space $Y$; by connectedness, $f(X) = Y$).

Answer 2

No. Consider the map $f : [0, 1] \to [0, 1]$ given by $f(x) = 0$. Note that $\partial[0, 1] = \{0, 1\}$ and $f(\{0, 1\}) = \{0\} \subset \{0, 1\}$. So $f$ need not be surjective nor must $f(\partial X) = \partial Y$.