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

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.

• 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 $U \to f[U]$. Jun 13, 2018 at 20:32

$f: X \to Y$ 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 $A$.