$f:A \backslash U\to B \backslash f(U)$ is also homeomorphic? $f:A \to B$ a homeomorphic mapping from $A$ to $B$
$=> f:A \backslash U\to B \backslash f(U)$ is also homeomorphic, $U \subseteq A$.
Why does this implication hold?

Since $f$ is homeomorphic it isn't difficult to show that $\bar{f}$ (the quotient mapping) is bijective. To prove continuity we take an open set $\bar{O}$ in $B\setminus f(U)$, which is of the form $O\cap f(U)$ with $O$ open in $B$. Now $\bar{f}^{-1}(\bar{O}) = f^{-1}(O)\cap A\setminus U$. We know that $f$ is a homeomorphism, so $f^{-1}(O)$ is open in $A$. Then by definition of the quotient topology $\bar{f}^{-1}(\bar{O})$ is open in $A\setminus U$.
