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).
Best Answer
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.