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Give $f: A\rightarrow B$ is a closed , onto and continuous mapping. For $b\in B$, let $C$ be an open set in $A$ s.t $\ f^{-1}(b)\subset C$. Show that there exists open set $D$ in $B$ with $b\in D$ s.t $f^{-1}(D)\subset C$.

My attempt Since $f$ is an onto and continuous mapping, for every open set $C$ in $A$, $f(C)$ is an open set in $B$. Since $\ f^{-1}(b)\subset C$, $f(C)$ contains $b$ and is open. Since $f$ is closed, there must exists an open set $D\subset B$ such that $f(C)$ contains $D$. Thus $f^{-1}(D)\subset C$ (Q.E.D).

I'm not quite sure if this proof is correct, especially where I used the hypothesis $f$ is closed. Can someone please help with this problem?

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    $\begingroup$ Why exactly does $f$ map open sets onto open sets? (Continuity says that the preimage of open sets is open, not the other way round. Consider $x\mapsto x^2$, which maps $(-1, 1)$ onto $[0, 1)$). $\endgroup$ – Thomas May 1 '15 at 16:34
  • $\begingroup$ @Thomas: That's a mistake indeed...:( (I need $f$ to be bijective for that to be true, but I can't show that:P). Can you please help with the proof? $\endgroup$ – user177196 May 1 '15 at 16:54
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    $\begingroup$ Just notice, that $f(A\setminus C)$ is closed, because $A\setminus C$ and also $f$ are closed. $\endgroup$ – Leonhardt von M May 1 '15 at 17:30
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$f$ is a closed mapping, and so if $U \subset A$ is closed in $A$, $f(U)$ is closed in $B$.

For $b \in B$, there is an open $C \subset A$ s.t. $f^{-1}(b) \subset C$ since $f$ is onto ($A$ is such a set). Consider any such $C$.

$A\backslash C$ is closed, so $f(A\backslash C)$ is closed (as $f$ is a closed mapping).

That is to say, $B\backslash f(A\backslash C)$ is open in $B$, also $ b \in B\backslash f(A\backslash C)$. Further, $$\begin{align*} f^{-1}(B\backslash f(A\backslash C)) &= A \backslash f^{-1}(f(A\backslash C))\\ &\subset A \backslash (A \backslash C) \\&= C \end{align*}$$ Where the first line follows as $f$ is onto, and the second is from the fact that $A \backslash C \subset f^{-1}(f(A\backslash C))$

Thus take $D = B\backslash f(A\backslash C)$ and we get $b \in D$, $f^{-1}(D) \subset C$

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