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I am reading a book on covering maps in Bourbaki-style with the following definitions:

A map $f:X\to Y$ is separated if for every $x,x'$ in the same fiber, there exist open disjoint neighbourhoods of $x$ and $x'$.

A map $f:X\to Y$ is proper if every filter $\mathcal{F}$ on $X$ such that $f(\mathcal{F})$ converges to $y\in Y$, admits a cluster point $x\in f^{-1}(y)$.

Given $f\colon X\to Y$, $g\colon Y\to Z$ such that $g\circ f$ is proper and $g$ is separated. Prove that $f$ is proper.

Let $\mathcal{F}$ be a filter on $X$ such that $f(\mathcal{F})\to y$, then $gf(\mathcal{F})\to g(y)$, so since $gf$ is proper, there exists a cluster point in $f^{-1}(g^{-1}((g(y)))$. However, I need to prove there exists a cluster point in $f^{-1}(y)$. I need to use separatedness of $g$ here somewhere, but I am new to filters and I do not see how I can use this here.

Any clues? Any help is appreciated, thank you!

Dr. Heinz Doofenshmirtz
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We have only to show that every cluster point $x\in(gf)^{-1}(g(y))$ of $ \mathcal F$ is also in $f^{-1}(y)$. Suppose not, then $f(x)\ne y$ are in the same fiber of $g$. By separatedness the two points have disjoint open neighborhoods $U\ni f(x)$ and $V\ni y$. By the assumption $f(\mathcal F)\to y$ and the definition of $f(\mathcal F)$ and filter convergence, there exists $F\in\mathcal F$ such that $f(F)\subseteq V$. To the contrary, one can find a point in $f(F)\setminus V$ using the fact that $U,V$ are disjoint and that $x$ is a cluster point of $\mathcal F$. Indeed, Since $x$ is a cluster point, it is in the closure of $F$ and we can choose an arbitrary point $x'\in f^{-1}(U)\cap F$. Then $f(x')\in f(F)$, but $f(x')\notin V$ since $f(x')\in U$.

Noiril
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