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Let there be an open surjective continuous function $f:X\to Y$, where $X$ and $Y$ are topological spaces. I am given to understand that this mapping need not be closed. But could you point out the flaw in the following proof?

Let the open set $A\subseteq X$ map to the open set $f(A)$. Also assume $f$ is surjective. Then the complement of $A$ or $A'$ maps to $f(A)'$. Both are closed as per the definition of closed sets.

I know counter-examples exist. I"m just looking to find the flaw.

Thanks in advance!

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Because of $f(A^C)\neq f(A)^C$ just think of constant functions. Here $A^C$ is the complement of $A$.

When your function is surjective you only know that

$f(A^C)\supseteq f(A)^C$ and equality holds when $f$ is bijective.

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  • $\begingroup$ I have just one further question. Say $U$ is an open set in $Y$. Then $f^{-1}(Y)$ is open in $X$. Can we say $(f(^{-1}(U))^C$ maps to $U^C$? In other words, can we say $f((f(^{-1}(U))^C)=U^{C}$? $\endgroup$ – fierydemon Aug 30 '13 at 8:21
  • $\begingroup$ @AyushKhaitan No that is only the case when $f$ is injective $\endgroup$ – Dominic Michaelis Aug 30 '13 at 8:24
  • $\begingroup$ Terribly sorry for another question: but this is throwing me off a little. Say $f:X\to Y$ is continuous. And $A,B$ and $C$ are open sets in $X$, and $f(A)=f(B)=f(C)=U$, where $U$ is an open set in $Y$. So what is $f^{-1}(U)$? Assuming no point in $X\setminus (A\cup B\cup C)$ maps to $U$, is $f^{-1}(U)=A\cup B\cup C$? $\endgroup$ – fierydemon Aug 30 '13 at 8:29
  • $\begingroup$ yeah you even don't need that any of $A,B,C$ is open $\endgroup$ – Dominic Michaelis Aug 30 '13 at 8:30
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    $\begingroup$ In that case, $f^{-1}(U)$ contains all points in $X$ that map to a point in $U$. Hence, $(f^{-1}(U))^C$ contains the points in $X$ which do not map to a point in $U$. They have to map to something. We also know that the function is surjective. Hence $U^C=f((f^{-1}(U))^C)$. Where am I going wrong? $\endgroup$ – fierydemon Aug 30 '13 at 8:41
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The problem is that $X\setminus A$ need not map to $Y\setminus f[A]$. Consider the map

$$\pi:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x$$

that projects the plane to the $x$-axis. This map is continuous and open. The set $L=\Bbb R\times\{0\}$ is closed, and $\pi[L]=\Bbb R$, but $\pi[\Bbb R^2\setminus L]=\Bbb R$ as well.

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  • $\begingroup$ but isn't $\mathbb{R}$ also closed so the closed set is still get to mapped to a closed set? $\endgroup$ – nekodesu Feb 12 '16 at 16:29
  • $\begingroup$ @Shiyue: The map is not closed: it sends the graph of $y=\frac1x$, which is closed in $\Bbb R^2$, to $\Bbb R\setminus\{0\}$, which is not closed in $\Bbb R$. But that’s beside the point: as requested, I was simply explaining what was wrong with Ayush’s reasoning. $\endgroup$ – Brian M. Scott Feb 12 '16 at 18:05

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