Let $f:X \to Y$ be a homeomorphism between topological spaces $X$ and $Y$. Let $A \subseteq X$ and $B = f(A) \subseteq Y$ (given the subspace topology) and let $g: A \to B$ be the restriction of $f$ given by $g(a) = f(a)$ for $a \in A$.
Showing that $g$ is a homeomorphism.
- $g$ injective
Let $g(a_1) = g(a_2)$, $a_1, a_2 \in A$.
$\implies f(a_1) = f(a_2)$
$\implies a_1 = a_2$ as $f$ homeomorphic.
- g surjective
Let $y \in B$
$\implies y = f(a)$ for some $a \in A$.
$\implies y = g(a)$ for some $a \in A$.
- $g$ continuous
Let $U \subseteq B$ be an open set in $B$.
$f(a) = g(a)$ for all $a \in A => f^{-1}(b) = g^{-1}(b)$ for all $b \in B$.
So $g^{-1}(U) = f^{-1}(U)$ which is open in $A$ as $f$ homeomorphic.
- $g^{-1}$ continuous
Let $U \subseteq A$ be an open set in $A$.
$g(U) = f(U)$ which is open in $B$ as $f$ homeomorphic.
So $g$ is homeomorphic.
Is my understanding correct, this seems almost too straightforward so I was wondering have I overlooked something?
Best Answer
In the part about continuity, note that $U\subseteq B$ open in $B$ means only that $U=B\cap U'$, where $U'$ is open in $Y$. Then $$g^{-1}(U)=f^{-1}(U)\cap A=f^{-1}(U')\cap f^{-1}(B)\cap A=f^{-1}(U')\cap A$$ so it is an open subset of $A$. This equation works as long as $f(A)⊆B$.
Or show the continuity of $g$ by the following diagram
$$\begin{array}{ccc} \ A & \xrightarrow{i} & X\\ g\downarrow & & \ \downarrow f\\ \ B & \xrightarrow{j} & Y \end{array}$$ By the universal property, $g$ is continuous if $jg$ is, but that's the same as $fi$ which is a composition of continuous functions.
You could also try to find the inverse of $g$. If $\bar f:Y→X$ denotes the inverse of $f$, then its restriction $\bar g$ to $B$ is a map $\bar g:B\to A$ since $\bar f(B)=f^{-1}(B)=f^{-1}(f(A))=A$. Now $$\bar gg(a)=\bar ff(a)=a\qquad g\bar g(b)=f\bar f(b)=b$$ So $g $ and $\bar g$ are inverse to each other.