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Is every open surjection also closed? This confuses me. It is mentioned that this is not true, but I can't see why.

Suppose that $f: X \to Y$ is an open surjection between topological spaces. Then for every closed $C \subset X$ the set $Y \setminus f(X \setminus C)$ is closed. Does the equality $Y \setminus f(X \setminus C) = f(C)$ hold? Let's see.

Let $y \in Y$ be some arbitrary element. Then $y \not\in f(X \setminus C)$ iff $f^{-1}(y) \cap (X \setminus C) = \varnothing$ iff $f^{-1}(y) \subset C$ iff $f^{-1}(y) = \varnothing$ or $y \in f(C)$, so for a surjection openness and closedness must be equivalent, right? Something must elude me here, but embarassingly I don't understand what it is.

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The graph of $y = 1/x$ ($x \neq 0$) is closed in $\mathbb{R}^2$ but its projection to the $x$-axis is $\mathbb{R} \smallsetminus \{0\}$. – t.b. Jun 4 '11 at 18:15
up vote 2 down vote accepted

The equality $Y\setminus f(X\setminus C)=f(C)$ doesn't hold if $f$ isn't a bijection. Suppose that $y=f(a)=f(b)$ with $a\in C$, $b\not\in C$. Then $y\in f(X\setminus C)$, so $y\not\in Y\setminus f(X\setminus C)$, but $y\in f(C)$.

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The projection $p: X\times Y \to X$ is open (and surjective) for any topological spaces $X$, $Y$. See e.g.

But in general it is not closed. Consider $\mathbb R\times\mathbb R$ and the subset $V=\{(x,y); y\ge \frac 1x\}$. The set $V$ is closed, but $p[V]$ is not.

By the way, your argument works for bijections:

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In fact, said projection $p$ is only garantueed to be closed when $Y$ is compact. – Henno Brandsma Jun 5 '11 at 17:47
Thanks for pointing this out @Henno. This is quite a useful fact, known as Kuratowski theorem or Kuratowski-Mrowka theorem. – Martin Sleziak Jun 6 '11 at 12:18

The problem is in $f(C)=Y-f(X-C)$. You are assuming that passing to the image of a function preserves differences of sets, which is false: take $f(x)=x^2$, with sets $[-1, 1]$ and $[0, 1]$: you have

$$f([-1, 1]-[0, 1])=(0, 1]$$ $$f([-1, 1])-f([0, 1])=\varnothing.$$

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