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).

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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.

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You're right, continuous functions do not suffice, is it always true when f is an homeomorphism? – Temitope.A Jul 14 '12 at 19:44
I should have be more specific, with distinguishing I meant if they were homeomorphic or not. If the boundary of two sets are not homeomorphic, then the sets themselves are not, while homeomorphism between boundaries, of course, does not suffice to make the sets homeomorphic. (?) – Temitope.A Jul 14 '12 at 19:50
@Temitope.A: I've updated my answer. – Zev Chonoles Jul 14 '12 at 20:03
Oh, well! An example was given us in class, of a closed disk not beign homeomorfic to a closed-annulus because the boundaries were not. Is it because in this case the boundaries are part of the sets, or did I missheard completely? And, finally, if it were an open disk and an open annulus, which concepts of general topology could be used to disprove homeomorphism? Thank you. – Temitope.A Jul 14 '12 at 20:17
@Temitope.A: Yes, it’s because the boundaries are part of the sets, and they can be distinguished from the rest of the sets: they are the points that don’t have open nbhds (in the set) homeomorphic to $\Bbb R^2$. The open annulus can be distinguished from the open disk by the fact that it has a two-point compactification, and the open disk does not. – Brian M. Scott Jul 14 '12 at 21:49