In trying to prove that the graph of a continuous map of compact Hausdorff spaces, $f:X\to Y$ is compact, I stumbled on this problem:
Let $f:X\to Y$ be a continuous function, $U$ and $V$ open in $X$ and $Y$ respectively. Let $A=f(U)\cap V$. Then $U \cap f^{-1}(A)$ is open in $X$.
Under what conditions is this true, and can a proof be provided.
Best Answer
First, $f(U)$ not necessarily is open, e.g., let $f:\mathbb{R}\to\mathbb{R}$ with $f(x)=x^2$ is continuous, but $f(\mathbb{R})=[0,+\infty)$ is not open in $\mathbb{R}$.
Sencond, $U\cap f^{-1}(V)$ is open, since $f^{-1}(V)$ is open (since $f$ is continuous) and finite intersection of open is open.
Fnally, $U\cap f^{-1}(A)=U\cap f^{-1}(V)$. Since $f^{-1}(A)\subset f^{-1}(V)$, then $U\cap f^{-1}(A)\subset U\cap f^{-1}(V)$. If $x\in U\cap f^{-1}(V)$, then $f(x)\in V$ and $f(x)\in f(U)$, i.e., $f(x)\in A$. Thus, $$x\in f^{-1}(f(x))\subset f^{-1}(A)\;,$$ so $x\in U\cap f^{-1}(A)$, and this implies that $U\cap f^{-1}(V)\subset U\cap f^{-1}(A)$, showing that $$ U\cap f^{-1}(A)=U\cap f^{-1}(V). $$ Therefore, $U\cap f^{-1}(A)$ is open.