If $f$ is a homeomorphism and $f(\partial B)=\partial B$, can we conclude $f(B^\circ)=B^\circ$ and $f(\overline B)=\overline B$?

Question

Let $E_i$ be a topological space, $f:E_1\to E_2$ be a homeomorphism and $B_1\subseteq E_1$.

We can easily show that

1. $f(B_1^\circ)={f(B_1)}^\circ;$
2. $f(\partial B_1)=\partial f(B_1)$; and
3. $f(\overline{B_1})=f(B_1^\circ)\cup f(\partial B_1)=\overline{f(B_1)}$.

Now assume $E_1=E_2$ and $f(\partial B_1)=\partial B_1$. Can we conclude that

4. $f(B_1^\circ)=B_1^\circ$; and
5. $f(\overline{B_1})=\overline{B_1}$?

If necessary, feel free to assume that $E_1$ is metrizable.

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Intuitively, both claims seem to be obvious, but I don't know how we can prove them. For (4.), maybe we need to take $x:=f(B_1^\circ)$ and assume $y:=f(x)\not\in B_1^\circ$. From $(1.)$, we know that $y$ is in the interior of $f(B_1)$. Maybe we can show that this is a contradiction ...

Answer 1

If $E_1=E_2=[0,1]$, $B=[0,\frac 1 2] \cup \{1\}$ and $f(x)=1-x$ the $f$ is a homeomorphism and $f(\partial B)=\partial B$. But $f$ does not map the interior of $B$ to interior of $B$ or the closure of $B$ to the closure of $B$.

Answer 2

I admit I haven't checked the details, but you should get a counterexample by letting $B \subseteq \mathbb{C}^*$ be the unit ball in the space of nonzero complex numbers and considering the inversion map $\mathbb{C}^* \to \mathbb{C}^*$, $z \mapsto z^{-1}$.

Answer 3

If you consider $B_1$ and $f(B_1)$ as topological spaces equipped with the subspace topology then you clearly have that $f(B_1^{\circ}) \cong B_1^{\circ}$ and $f(\overline{B_1})\cong\overline{B_1}$, but not equality in general.

Here's a geometrical counterexample. Take $E_1$ to be the closed disk of radius $1$ centered at the origin in $\mathbb R^2$, $B_1$ to be the closed disk of radius $1/4$ centered at $(1/2, 0)$ in $\mathbb R^2$ and $f :E_1 \to E_1$ be the homeomorphism given by the anticlockwise rotation of $\mathbb R^2$ around the origin on an angle $\pi$.