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Suppose that $f : X \to Y$ is a homeomorphism and $U$ is an open subset of $X$. Show that $f[U]$ is open in $Y$ and the restriction $f|_U$ is a homeomorphism from $U$ to $f[U]$

I showed that $f[U]$ is open in $Y$. I'm trying to show that the restriction $f|_U$ is a homeomorphism from $U$ to $f[U]$

There are two theorems I'll use to prove this.

Let $X$ and $Y$ be topological spaces

Theorem 1: If $f : X \to Y$ is continuous and if $A$ is a subspace of $X$, then the restricted function $f|_{A}: A \to Y$ is continuous

Theorem 2: $f : X \to Y$ is continuous if and only if $f : X \to f[X]$ is continuous.

The two above theorems combined give the following lemma

Lemma: Let $X$ and $Y$ be topological spaces and let $A$ be a subspace of $X$. If $f : X\to Y$ is continuous then the restricted function $f|_{A} : A \to f[A]$ is continuous.

Now onto my attempted proof,

My Attempted Proof: Since $f : X \to Y$ is a homeomorphism we have that $f$ is bijective and continuous and it's inverse $f^{-1} : Y \to X$ is continuous.

Let $f|_{U} : U \to f[U]$ be the restriction of $f$ to $U$. Trivially we have that $f|_U$ is bijective since $f$ is bijective. Since $f$ is continuous then by the above lemma since $U$ is a subspace of $X$ we have that $f|_{U}$ to be continuous.

Similarly let $\left(f|_{U}\right)^{-1} : f[U] \to U$ denote the inverse of $f|_U$ then again by the above lemma since $f[U]$ is a subspace of $Y$ and since $f^{-1}$ is continuous we have that $\left(f|_{U}\right)^{-1}$ is continuous. Hence $f|_U$ is a homeomorphism. $\square$


First of all is my proof correct? If it is then why do we need $U$ to be an open subset of $X$ for the restriction to be a homeomorphism in the first place? I haven't used the fact that $U$ is an open subset of $X$ anywhere in my proof, in fact my proof as far as I can tell shows that the restriction of $f$ to any subspace $A \subseteq X$ yields a hoemomorphism between $A$ and $f[A]$, thus either my proof is incorrect or what I've just said must hold.

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  • $\begingroup$ Your argument that $\bigl(f\rvert_U\bigr)^{-1}$ is continuous doesn't really follow from the lemma. The lemma gives you the continuity of $\bigl(f^{-1}\bigr)\rvert_{f[U]}$, and then the observation that $\bigl(f\rvert_U\bigr)^{-1} = \bigl(f^{-1}\bigr)\rvert_{f[U]}$ takes care of the rest. And yes, one doesn't need that $U$ is open to conclude that $f$ induces a homeomorphism $U \to f[U]$. $\endgroup$ Jun 13, 2018 at 20:32

1 Answer 1

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The proof is essentially correct: you need that restrictions (on both image and domain side, when defined) of continuous functions are continuous. Plus some simple set theory facts on inverse functions, formally:

$f: X \to Y$ continuous, so $f|_A: A \to Y$ continuous (domain restriction), and hence $(f|_A)': A \to f[A]$ continuous (codomain restriction).

If $g: Y \to X$ is the continuous inverse of $A$, $g|_{f[A]} \to X$ is continuous (domain restriction) and so $(g|_{f[A]})': f[A] \to g[f[A]]$ is continuous (codomain restriction).

We know that $g \circ f = 1_X$ and $f \circ g = 1_Y$, so in particular $g[f[A]] = A$ and $(g|_{f[A]})'$ is the inverse of $(f|_A)'$: domains and codomains match and the same two equations still hold: $$(g|_{f[A]})' \circ (f|_A)' = 1_A$$ and $$ (f|_A)' \circ (g|_{f[A]})' = 1_{f[A]}\text{.}$$ So indeed $(f|_A)': A \to f[A]$ is a homeomorphism as well.

Neither openness nor closedness of $A$ are relevant. It holds for all $A$.

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