# Homeomorphism between topological subspaces

Let $$X$$ be a topological space. Take $$A,B\subseteq X$$ and assume that there is homeomorphism $$\phi:A\rightarrow B$$.

Is it true that $$\phi(int(A))=int(B)$$?

I know that homeomorphism preserve all topological structure so it seems true. But $$int(A)$$ can't be defined by the topology of $$A$$ alone, it must use the topology of $$X$$. Hence it sound reasonable to say that the homemorphism dosn't surely take the interior to the interior.

This being said, it sound strange to me that we have, for example, a homemorphism between some cube and some ball but in some way it doesn't map interior to interior. Do you have an exmplae for that?

Here is a counterexample:

$$X = [0,1]$$, $$A = [0,1]$$, $$B = [0,1/2]$$, $$h(a) = a/2$$.

Many counterexamples exist in abstract metric spaces, one of them given by Paul Frost. But given your particular objection:

It sound strange to me that we have, for example, a homemorphism between some cube and some ball but in some way it doesn't map interior to interior.

it could be interesting for you that the theorem holds when $$X = \mathbb{R}^n$$ with the standard topology. That is,

Assume $$A, B \subseteq \mathbb{R}^n$$ and $$\phi : A \to B$$ is a homeomorphism. Then $$\phi[ \operatorname{int} A ] = \operatorname{int} B$$.

To prove it, we can use the domain invariance theorem:

If $$U \subseteq \mathbb{R}^n$$ is open and $$f : U \to \mathbb{R}^n$$ is continuous and injective, then $$f[U]$$ is open.

From the theorem we quickly get that $$\phi[\operatorname{int} A] \subseteq B$$ is open in $$\mathbb{R}^n$$, so $$\phi[\operatorname{int} A] \subseteq \operatorname{int} B$$. The other inclusion follows by symmetry. $$\square$$

So indeed, a homeomorphism between a cube and a ball must preserve the interior.