A homeomorphism maps boundary in boundary? Let $U$ and $V$ be two topological spaces and let $A \subset U$ be a topological subspace of $U$ and let $B \subset V$ be a topological subspace of $V$.
Let $f:A \to B$ be a homeomorphism 
I would like to know if is it true that
$$
a \in \delta A \Rightarrow f(a) \in \delta B
$$
($\delta X =$ boundary $X$)
 A: This is an example in which you won't have the problem of well definiteness of  other answers. Take 
$U=\mathbb{R}^2$ with the usual topology,
$V= \mathbb{R}$ with the usual topology,
$A=\{(x,y)\in \mathbb{R}^2:y=0\}$,
$B=V,$ 
$f:A\to B$ as $f(x,0)=x.$ 
A: In general, you don't even know if $f(a)$ is defined, since you do not know if $\delta A\subset A$.
For example, take $U=V=\mathbb R$, $A=B=(0,1)$ (standard topology, of course), and $f:A\to B$ defined as $f(x)=x$.
Then, $0\in\delta A$ and $f(0)$ is not defined.
A: In general $f$ does not extend to $\partial A\setminus A$ so the question does not make much sense.
For instance, consider $A=(-\frac\pi2,\frac\pi2)\subset U=\Bbb R$ and $B=V=\Bbb R^2$ with $f(x)=(x,\tan x)$.
If $f$ is the restriction of some continuous function $\tilde f$ defined on some neighborhood containing $\bar A$ then certainly $f(\partial A)\subseteq\bar B$.
A: The following statements are equivalent, given a map $f: A\rightarrow B$, $A\subset U$ and $B\subset V$ topological spaces:


*

*the map is continuous

*the preimage of the family of the open sets in $V$ is contained in the family of the open sets in $U$.

*the preimage of the family of the neighborhoods of a point is contained in the family of the neighborhoods of any preimage of the point.

*Being a boundary point of a set for a point and a set in the domain of the map is an invariant property under the map.


Hope this help. Note also that homeomorphism is not needed.

the fourth point means:
4'. Given a point and a set both in the domain of the map, if the point is a boundary point of the set the image of the point is a boundary point of the image of the set under the map
