Q:

How does homeomorphism map sets boundaries?

I'm at the end of my first course on general topology, but this topic was not well developed.
I can tell that an homeomorphism preserves the quality of a point to be a boundary point for a subset of a topological space. In particular, from space X to space Y, one only needs a function to be countinuos (or something else? I think there's not even need for bijectivness).
But what happens when we talk about The boundary as a whole? What happens in terms of connected component et cetera?
Am I right to say we can distinguish two subspaces by their boundaries even when this is not included in the subspace? (Specially when considering the topology on an open subset).

A:

I'm not sure I understand your question, but perhaps this example will help: 
Consider the continuous function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^2$. Let our set be $S=[-2,1]$. Then $\partial S=\{-2,1\}$, but the image of $S$ is $f(S)=[0,4]$, which has boundary $\partial f(S)=\{0,4\}$, so that $f(1)$ is not in the boundary of $f(S)$ even though $1$ was in the boundary of $S$. 
Regarding "distinguishing two subspaces by their boundaries": let $X=\{a,b,c\}$ with the topology $T=\{\varnothing,\{a\},\{b\},\{a,b\},X\}$. Let $A=\{a\}$ and $B=\{b\}$. Then 
$$\partial A=\overline{A}\setminus A^o=\{a,c\}\setminus \{a\}=\{c\}\quad\text{and}\quad \partial B=\overline{B}\setminus B^o=\{b,c\}\setminus\{b\}=\{c\}$$
so that $\partial A=\partial B$, even though $A\neq B$, and both $A$ and $B$ are open subsets of $X$ that do not contain their boundaries.

Consider the unit circle $\mathbb{S}^1=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2=1\}$. Let $A,B\subseteq\mathbb{S}^1$ be
$$A=\{(\cos(2\pi t),\sin(2\pi t))\mid t\in (0,\tfrac{1}{2})\}\quad\text{ and }\quad B=\{(\cos(2\pi t),\sin(2\pi t))\mid t\in (0,1)\}.$$
Then $A$ and $B$ are homeomorphic to each other, even though 
$$\partial A=\{(1,0),(-1,0)\}\quad\text{ and }\quad \partial B=\{(1,0)\}$$
are not homeomorphic to each other.

