Do (actual) homeomorphisms preserve closures of subspaces?

Question

My question is

Do (actual) homeomorphisms preserve closures of subspaces?

There's another question

Do homeomorphisms preserve closures of subspaces?

Do homeomorphisms preserve closures of subspaces? ( I think duplicate of "closure preserves homeomorphism" )

But I think title is wrong because the ambient spaces aren't homeomorphic. In that question we have topological spaces $X$ & $Y$ which are unrelated except that they have respective subspaces $A, B$ that happen to be homeomorphic $f: A \to B$. Expectedly, the closures of $A$ and $B$ are not necessarily homeomorphic, just like $A=(0,1)$ and $X=Y=B=\mathbb R$.

This time, I want to know what happens if the homeomorphism $f$ comes from a homeomorphism between the ambient topological spaces:

Let $g: X \to Y$ be a homeomorphism. Suppose $g$'s restriction to $A$ gives a homeomorphism $f: A \to B$. Explicitly:

1. Restrict domain of $g$ to $A$ to get the map $g|_A : A \to Y, (g|_A)(x) := g(x)$.
2. Suppose $image(g)=g(A)=B$.
3. Now restrict range of $g|_A$ to $B$ to get the map $f:A \to B, f(x) = g(x)$.
4. Since $g$ is a homeomorphism, $g|_A$ and $f$ are injective.
5. Suppose $f$ is a homeomorphism too. (i.e. $g|_A$ is an embedding right?)

Hence, like in the previous question, we have a homeomorphism $f: A \to B$ between the subspaces $A$ & $B$. But now I'm additionally supposing this $f$ is a homeomorphism derived from a homeomorphism $g: X \to Y$ between the respective ambient topological spaces $X$ and $Y$.

If $g$ is not a homeomorphism, I don't see why $\text{closure}_X(A) = \text{closure}_Y(B)$. But now that $g$ is, maybe now $g(\text{closure}_X(A)) = \text{closure}_Y(B)$...

...Actually, apparently not. According to this Why are the closures of $(0,1)$ and $(0,+\infty)$ not homeomorphic despite the intervals themselves being homeomorphic?

$$A=(0,1), B=(0,\infty), X = Y = \mathbb R, \ \text{then whatever f & g you like}$$

Questions:

1. Ok so generally, when do (actual) homeomorphisms preserve closures of subspaces, like for a subspace $C \subseteq X$, when do we have $g(\text{closure}_X(C)) = \text{closure}_Y(g(C))$, assuming $C$ & $g(C)$ are homeomorphic?

2. Btw, do homeomorphisms always restrict to homeomorphisms? Eg in above are $C$ & $g(C)$ homeomorphic (and hence the assumption is redundant)?

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Edit to add: Ah wait based on these 2 answers:

1. Why are the closures of $(0,1)$ and $(0,+\infty)$ not homeomorphic despite the intervals themselves being homeomorphic?

2. Homeomorphism that maps a closed set to an open set?

It seems

$$g(\text{closure}_X(C)) = \text{closure}_Y(g(C))$$

is always true whether or not $C$ & $g(C)$ are homeomorphic and then

$$A=(0,1), B=(0,\infty), X = Y = \mathbb R, \ \text{then whatever f & g you like}$$

isn't a counter-example because actually no such $f,g$ exist, i.e. there isn't an automorphism $f:\mathbb R \to \mathbb R$ with $f(0,1)= (0,\infty)$

Answer 1

Homeomorphisms always preserve closure of subspaces. If $f:X\to Y$ is a homeomorphism, $A\subseteq X, B\subseteq Y$ and $f(A)=B$, then $f(\overline{A})=\overline{B}$. But when I say closure, I mean the closure of $A$ as a subspace of $X$. So if you take $X=A=(0,1)$ then you have to consider the closure of $(0,1)$ as a subspace of $(0,1)$, not as a subspace of $\mathbb{R}$. And the closure of $(0,1)$ as a subspace of itself is the set $(0,1)$ itself. So your example is not a real counterexample.

Here is one possible way to prove that in the above notations indeed $f(\overline{A})=\overline{B}$. Take $x\in\overline{A}$. We will show that $f(x)\in\overline{B}$. Let $U\subseteq Y$ be an open neighborhood of $f(x)$, we have to show it intersects $B$. By continuity, $f^{-1}(U)$ is an open neighborhood of $x$ in $X$, hence it contains some element $a\in A$. But then $f(a)$ is an element of $U\cap B$, so $U$ indeed intersects $B$. This shows that $f(\overline{A})\subseteq\overline{B}$. The other inclusion is symmetric, as $A=f^{-1}(B)$, and $f^{-1}$ is also continuous.

As for the other question, yes, a restriction of a homeomorphism is always a homeomorphism. This is almost immediate from the restriction of a continuous map being continuous.

Answer 2

Yes, given a homeomorphism $f:X\to Y$, a subset $A\subseteq X$, and its image $B=f(A)\subseteq Y$, we have indeed $$f(\overline A)=\overline B,$$ and here is the simplest proof (imo). It relies on the characterization of $\overline A$ (in $X$) as the smallest closed set containing $A$.

By continuity of $f$, the set $f^{-1}(\overline B)$ is closed. Moreover, it contains $f^{-1}(B)=A$. Therefore, $f^{-1}(\overline B)\supseteq\overline A$, i.e. $$\overline B\supseteq f(\overline A).$$ Similarly, by continuity of $g:=f^{-1}$: $g^{-1} (\overline A)\supseteq\overline B$, i.e. $$f (\overline A)\supseteq\overline B,$$ which ends the proof.