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Let $X$ and $Y$ Hausdorff spaces and $f:X\rightarrow Y$ a function. Prove that the next sentences are equivalent.

1) $f$ is a perfect function

2) For all $y\in Y$, the set $f^{-1}[\{y\}]$ is closed in $X$ and for all $\mathcal{U}$ open cover of $X$, closed under finite unions, i.e., if $\left\{U_1,...,U_n \right\}\subseteq \mathcal{U}$ then $\displaystyle\bigcup_{i=1}^{n}U_i\in \mathcal{U}$, then the set $$\mathcal{V}=\left\{ Y\setminus f[X\setminus U] \ | U\in\mathcal{U}\right\}$$is an open cover of $Y$.

First, the definition of a perfect function

Definition: $f:X\rightarrow Y $ is a perfect function if $f$ is closed and if for all $y\in Y$, $f^{-1}[{y}]$ is a compact set

PROOF

1)$\Rightarrow 2)$

Let $y\in Y$, then, because $f$ is a perfect function, then $f^{-1}[\{y\}]$ is compact in a Hausdorff space. Thus, $f^{-1}[\{y\}]$ is closed.

Let $\mathcal{U}$ an open cover of $X$ closed under finite unions. Is clear that for all $U\in\mathcal{U}$, $U$ is open. Then, $X\setminus U$ is closed. Because $f$ is closed, then $f[X\setminus U]$ is closed and so, $Y\setminus f[X\setminus U]$ is open. Finally, every element of $\mathcal{V}$ is open.

Now, we want to prove that $Y=\bigcup\mathcal{V}$. Clearly, $\bigcup\mathcal{V}\subseteq Y$. Let $y\in Y$, then, $f^{-1}[\{y\}]$ is compact because the function is perfect. In this way, $\mathcal{U}$ is an open cover of $f^{-1}[\{y\}]$. Thus, there exist $\mathcal{U}_0\subseteq\mathcal{U}$ a finite subcover such that $f^{-1}[\{y\}]\subseteq \bigcup \mathcal{U}_0$. Because $\mathcal{U}$ is closed under finite unions, then, $\bigcup\mathcal{U}_0\in \mathcal{U}$. Then, $f^{-1}[\{y\}]\cap (X\setminus \bigcup\mathcal{U}_0)=\emptyset$, and $y\notin f[X\setminus\bigcup\mathcal{U}_0]$ and finally, $y\in Y\setminus f[X\setminus\bigcup\mathcal{U}_0]$. Thus, $Y=\bigcup\mathcal{V}$

2) $\Rightarrow$ 1)

First, we will to prove that $f$ is a closed function.

Let $K\subseteq X$ closed set. Then, $X\setminus K$ is open. Then, the set $\mathcal{U}=\{X\setminus K, X \}$ is an open cover of $X$ closed under finite unions. Then, by hiphotesys, the set $\mathcal{V}=\{ Y\setminus f[X\setminus(X\setminus K)],Y\setminus f[X\setminus X]\}$ is an open cover of $Y$. But, $Y\setminus f[X\setminus(X\setminus K)]=Y\setminus f[K]$ is open, then, $f[K]$ is closed. Thus, $f$ is closed.

Here is the problem. I don't have any idea of how can I prove the compactness of $f^{-1}[\{y\}]$. I think in the contradiction, but, really, I don't know in which moment should be used the open cover closed under finite unions. Any hint?

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Let $y \in Y$ and let $\mathcal{U}$ be some open cover of $f^{-1}(\{y\})$. Let $\mathcal{U}'$ be the set of all finite unions of elements of $\mathcal{U}$ and $f^{-1}(\{y\}^c)$, which is an open cover of $X$. If there is $n \in \mathbb{N}$ and $U_{i}' \in \mathcal{U}'$ with $\cup_{i=1}^{n}U_{i}' \supseteq f^{-1}(\{y\}), $ we are done because this implies that there are finitely many $U_j \in \mathcal{U}$ that cover $f^{-1}(\{y\})$.

Assume that for all $n \in \mathbb{N}$, this is not the case, implying that for all $U' \in \mathcal{U}'$, we have $f^{-1}(\{y\})\cap (U')^c \neq \emptyset$. As a consequence, $y \in f(X\setminus U')$ for all $U' \in \mathcal{U}'$ hence $y \notin Y\setminus(f(X\setminus U'))$ for all $U' \in \mathcal{U}'$, contradicting the fact that $\mathcal{V}$ is a cover of $Y$.

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