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I've proven that $Y$ is $T_1$ using $f$ proper $\Rightarrow $ $f$ closed and since X is Hausdorff, X is $T_1$, then the unitary sets are closed in $X$.

$f$ is quotient so it's surjective, then for all $y \in Y$ exists $x \in X$ that $f(x) = y $.

$\{x\}$ is closed $\Rightarrow \{y\}$ is closed for all $y\in Y$, then $Y$ is $T_1$ but I can't prove the Hausdorff condition.

Additional information: $f: X\to Y$ is $\textbf{proper}$ if for all $Z$ topological space, $g: Z \to Y$ continuous function, the projection $p: X \times_Y Z \to Z$ is closed.

$X \times_Y Z = \{(x,z)\in X\times Z : f(x) = g(z)\}$

Martin Sleziak
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Christian Coronel
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  • I don't have a solution for you but I imagine it helps to use that $Y$ is Hausdorff if and only if the diagonal $\Delta_Y$ is closed in $Y\times Y$ – TY Mathers Nov 07 '23 at 20:07
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    What is your definition of proper function? – freakish Nov 07 '23 at 20:43
  • @freakish $f: X\to Y$ is proper if for all $Z$ topological space, $g: Z \to Y$ continuous function, the projection $p: X \times_Y Z \to Z$ is closed. $X \times_Y Z = {(x,z)\in X\times Z : f(x) = g(z)}$ – Christian Coronel Nov 08 '23 at 03:32
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    I can finish the proof using another definition of proper map that $f$ is continuous and closed with compact fibers, however I'm struggling to convert your definition to this definition. – Ricky Nov 08 '23 at 03:45

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First, we need the following equivalent definition of a proper map (I don't have enough knowledge of algebraic topology to understand its proof but see the link).

Lemma 1 Given a continuous map $f:X\to Y$, the following properties are equivalent:
(1)$f$ is a closed map with compact fibers.
(2)$f$ is universally closed, that is, the pullback map $p: X \times _ Y Z \to Z$ is closed for any continuous map $g: Z \to Y$.

Lemma 2 If $f:X\to Y$ is closed, $y\in Y $, and $U$ is an open subset of $X$ such that $f^{-1}({y})\subseteq U $, then there is an open neighborhood $V$ of $y$ such that $f^{-1}(V)\subseteq U $.
Proof Consider $V=Y\setminus f(X\setminus U) $, which is open in $Y$. Since $f^{-1}({y})\subseteq U $, $y\notin f(X\setminus U) $, thus $y\in V $. Moreover, for every $z\in V $, $z\notin f(X\setminus U)$, hence $f^{-1}(z)\subseteq U .$

Theorem Let $f:X\to Y$ be a closed surjective map with compact fibers, if $X$ is Hausdorff, $Y$ is also Hausdorff.

Proof Let $a,b\in Y,a\ne b$, then $f^{-1}(a)$ and $f^{-1}(b)$ are disjoint nonempty compact subspaces of a Hausdorff space $X$, thus they can be separated by open neighborhoods $U\supseteq f^{-1}(a),V\supseteq f^{-1}(b)$. Now since $f$ is closed, there are open neighborhoods $A$ of $a$ and $B$ of $b$ such that $f^{-1}(A)\subseteq U$ and $f^{-1}(B)\subseteq V$. Since $U$ and $V$ are disjoint, so are $A$ and $B$. Therefore, $Y$ is Hausdorff.

Ricky
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