I am self studying Topology, and I am having difficulty in proving the following statement.
Let $f$ be a continuous mapping defined on a compact space $X$ onto a Hausdorff space $Y$. Suppose $y\in Y$ and $U$ is an open subset of $X$ that contains $f^{-1}[y]$. Show that there exists an open neighborhood $V$ of $y$ such that $f^{-1}[V] \subset U$.
I know that
- $Y=f[X]$ is compact, and thus, being Hausdorff, it is also normal;
- $f$ is a closed map;
- $f^{-1}[y]$ is closed and compact.
Can you give me some hints on how to use these facts (or other consequences of the hypothesis) to prove the theorem?
Best Answer
Because $U$ is open $X\backslash U$ is closed and therefore compact. Then $f(X\backslash U)$ is closed because $Y$ is a Hausdorff space. Also $y\not\in f(X\backslash U)$ because $f^{-1}(y)\subset U\ \cap X\backslash U = \varnothing$.
Let denote $V=Y\backslash f(X\backslash U)$ it is open and $y\in V$. $$f^{-1}(V)=f^{-1}(Y)\backslash f^{-1}(f(X\backslash U)) = X\backslash f^{-1}(f(X\backslash U))=W$$
$W$ is open because $f$ is continuous and $f(X\backslash U)$ is closed.
$X\backslash U \subset f^{-1}(f(X\backslash U))$ and therefore $W = X\backslash f^{-1}(f(X\backslash U)) \subset X\backslash(X\backslash U) = U$
$W\subset U$ as we need.