Do homeomorphic manifolds with boundary have homeomorphic interiors?

Question

Let $M, N$ be manifolds with boundary and $f: M \rightarrow N$ be a homeomorphism. I want to show that $\text{Int}M$ is homeomorphic to $\text{Int}N$. I think I have most of the proof but it relies on an assumption which I don't know if it's valid or not.

I first assume that $f(\text{Int}M) = \text{Int}N$. I don't have a proof of this, and as far as I know there could be counterexamples. But with this assumption the rest of the proof becomes very easy since f is the homeomorphism:

$f$ restricted to $\text{Int}M$ will clearly still be bijective, and a set is open in $M$ iff it is open in $\text{Int}M$. Same for $N$ and $\text{Int}N$. Thus continuity of $f$ and its inverse follow immediately, and so $\text{Int}M$ and $\text{Int}N$ are homeomorphic.

What about the assumption I made? Are there any counterexamples?

Your assumption is valid. The proof for topological manifolds I know of uses algebraic topology. Locally at interior points the punctured neighborhoods are spheres. Locally at boundary points, the puncture neighborhoods are disks. — jgon, Jul 26 '19 at 17:01
So it's always true that a homeomorphism maps interior to interior? — Paul Cusson, Jul 26 '19 at 17:54
The fact that $f(\operatorname{int}(M))=\operatorname{int}N$ for topological manifold homeomorphisms follows from the invariance of domain, which, as @jgon mentioned, uses algebraic topology. For topologial manifolds and $C^1$ diffeomorphisms you can use the inverse function theorem. — Luiz Cordeiro, Jul 27 '19 at 02:05

score 3 · Accepted Answer · answered Jul 29 '19 at 14:23

The fact that $f(\operatorname{int} M )=\operatorname{int} N$ for topological manifold homeomorphisms follows from the invariance of domain, which, as @jgon mentioned in the comments, uses tools from Algebraic Topology. See Elementary proof of topological invariance of dimension using Brouwer's fixed point and invariance of domain theorems? for good references.

Here are some details: Let $x\in\operatorname{int}M$. Taking small coordinate charts around $x$ and $f(x)$, and composing with the homeomorphism $f$, we find a homeomorphism $f'$ between an open set $U$ of $\mathbb{R}^n$, and an open subset $V$ of some $\mathbb{R}^m$, or an open subset $V$ of a half-space of $\mathbb{R}^m$. Restricting (the inverse of) this homeomorphism to the interior of $V$ (in $\mathbb{R}^m$), we see, by invariance of domain, that $m=n$ (see accepted answer to linked question above). But then invariance of domain itself implies that $V$ is open in $\mathbb{R}^n$, so $f(x)$ is interior to $N$.

This proves $f(\operatorname{int}M)\subseteq\operatorname{int}N$. The same argument with the inverse $f^{-1}$ gives the reverse inclusion.

The same type of argument works with $C^1$ diffeomorphisms between $C^1$ manifolds, using the inverse function theorem, which can be proven with relatively elementary real analysis and linear algebra.

Do homeomorphic manifolds with boundary have homeomorphic interiors?

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