Let $\phi: U \to V \subset \mathbb{R}^n$ homeomorphism.
My desire is:
I want to say the restriction $\phi|_{\phi^{-1}(B_{r'}(x))}:\phi^{-1}(B_{r'}(x)) \to B_{r'}(x) $ is a homeomorphism in the simplest way.
My understanding is:
Surjectivity) $\phi|_{\phi^{-1}(B_{r'}(x))} $ is surjective, since $\phi(\phi^{-1}(B_{r'}(x))) = B_{r'}(x)) $ (by surjectivity of $\phi$).
Injectivity) Clear.
Continuity of $\phi$) First, $\phi|_{\phi^{-1}(B_{r'}(x))}:\phi^{-1}(B_{r'}(x)) \to V $ is continuous,
since inclusion map is continuous and $\phi|_{\phi^{-1}(B_{r'}(x))} = \phi \circ \iota_{\phi^{-1}(B_{r'}(x))} $
Second, $\phi|_{\phi^{-1}(B_{r'}(x))}:\phi^{-1}(B_{r'}(x)) \to B_{r'}(x) $ is continuous,
since taking any $V'$:open in $B_{r'}(x)$, then there is some $W$: open in $B_{r'}(x)$ s.t. $V' = B_{r'}(x) \cap W$.
$\phi^{-1}( B_{r'}(x) \cap W) = \phi^{-1}( (\iota_{B_{r'}(x)})^{-1}(W))$, which is open in U.
Similarly, $\phi^{-1}$ is continuous.
I think this proof is valid, but can we show the continuity in more straightforward way, without breaking down to restricting the domain and codomain? Thanks.