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Suppose I have a continuous map $f:X\rightarrow Y$. Then one can wonder, whether for every open set $U\subset X$ the set $U':=\{x\in X|f^{-1}(f(x))\subset U\}$ is open again. This is not true in general, as the example

$U:=\{(x,y)|\quad |y|<\frac{1}{x^2+1}\}\subset \mathbb{R}^2$ and $f:\mathbb{R}^2\rightarrow \mathbb{R} \qquad (x,y)\mapsto y$

shows. Here $U'$ is just $\mathbb{R}\times \{0\}$. So the question is: Which continuous maps $f:X\rightarrow Y$ have the property, that for every open set $U\subset X$, the set $U'$ is also open?

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Currently I believe (due to a lack of counterexamples), that the right condition is "f is proper". –  HenrikRüping Mar 9 '10 at 16:00
    
Not quite so. For reasonable X and Y (e.g. locally compact Hausdorff) being proper is sufficient but not necessary. For example, every injective map has the property and there are non-proper injective maps. –  Sergei Ivanov Mar 9 '10 at 16:17
    
@Sergei: Why do X and Y need to be reasonable? –  François G. Dorais Mar 9 '10 at 16:52
    
Any example where some $f^{-1}(\{y\})$ is not compact? –  François G. Dorais Mar 9 '10 at 16:57
    
@François: you can construct any kind of weirdness when Y is anti-discrete (i.e., has no nontrivial open sets). Actually, I was thinking about metric spaces when I wrote "reasonable". –  Sergei Ivanov Mar 9 '10 at 17:12
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1 Answer

up vote 1 down vote accepted

The definition of $U'$ doesn't depend on the Topology on $Y$. Given any map $f:X\rightarrow Y$, we can consider the equivalence relation $x'\sim x$, iff $f(x)=f(x')$ on $X$ and factor the map $f$ as $X\rightarrow X/\sim \quad \rightarrow Y$. Call the first $f'$. Note that,the second map is injective. Hence we get the same set $U'$ for any set $U$, if we replace $f$ by $f'$.

So we might assume without restriction, that $f:X\rightarrow X/\sim $ is a quotient map. Then the upper statement is equivalent to saying, that $f$ is closed.

Suppose $f$ is closed. Then one gets for the complements $U'^c=f^{-1}(f(U^c))$. And hence $U'^c$ must be closed. There is the closed map lemma, which shows, that under the nice conditions mentioned in the comments above properness is also sufficient.

Now suppose $f:X\rightarrow X/\sim$ is not closed and a quotient map. Then there is a closed set $A\subset X$, such that its image is not closed. By definition of the quotient topology, a subset of $X/\sim$ is closed, if and only if its preimage is closed. So $f^{-1}(f(A))$ is not closed and hence $U=A^c$ gives the desired counterexample.

So we get: A map $f:X\rightarrow Y$ has the property, if and only if the induced map $X\rightarrow X/\sim$ is closed.

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Every injective map satisfies it. But not every injective map is closed. So e.g. the identity from [0,1] (usual topology) to [0,1] (cofinite topology) satisfies the condition, but is not closed. –  Henno Brandsma Mar 13 '10 at 6:30
    
@Henno: Ok I edited some details in the part, where I want to assume, that the map is a quotient map. In your example, the map is not a quotient map. But the map $f'$ is just the identity on $[0;1]$ and hence closed. –  HenrikRüping Mar 13 '10 at 9:40
    
Ok, I see. The statement was of the form: f has the property iff f' is closed. Still, one would like to see a more direct criterion. –  Henno Brandsma Mar 13 '10 at 13:05
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