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Let $X$ and $Y$ a topological space. If $f:X\rightarrow Y$ a surjective open map and $G_f = \{(x,f(x))\in X \times Y| x \in X\}$ is closed, then $Y$ is Hausdorff.

Can someone help me the process to prove the statement? I tried to prove it and obtain a two open sets of $Y$ but I'm stuck in showing that those are disjoint sets.

Edited: Here's my proof

Let $y_1, y_2 \in Y$. By surjectivity of $f$, there exist $x_1, x_2 \in X$ s.t. $f(x_1)=y_1$ and $ f(x_2)=y_2$.

Consider the point $(x_1,z_1)$ and $(x_2,z_2) \in X\times Y \setminus G_f$. Note that $z_1\neq f(x_1)$. Since $X\times Y \setminus G_f$ is open, there exist an open neighborhood $U_1,U_2$ of the points $(x_1,z_1)$ and $(x_2,z_2)$, respectively. We know that the projection map $\pi_1:X\times Y\rightarrow X $ is an open map and so, the set $\pi_1(U_1)$ and $\pi_1(U_2)$ is open in $X$. Furthermore, since $f$ is an open map, then the set $f(\pi_1(U_1))$ and $f(\pi_1(U_2))$ is open in $Y$. The set $f(\pi_1(U_1))$ and $f(\pi_1(U_2))$ are neighborhoods of $y_1$ and $y_2$ since $x_1 \in \pi_1(U_1)$ and $x_2 \in \pi_1(U_2)$.

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  • $\begingroup$ Can you be more specific about what you have tried? $\endgroup$ Jan 14, 2023 at 14:03
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    $\begingroup$ @C-RAM hello, i've edited my post to include my attempted proof $\endgroup$
    – yaechan
    Jan 14, 2023 at 14:21

2 Answers 2

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Let $y_1,y_2$ be distinct points in $Y$ and choose respective preimages $x_1,x_2$. Clearly, $(x_1,y_2)\not\in G_f$, so we can choose an open neighborhood of $(x_1,y_2)$ of the form $U\times V$ (with $U$ and $V$ open neighborhoods of $x_1\in X$ and $y_2\in Y$ respectiely), which is disjoint from $G_f$, by the definition of product topology (and since the complement of $G_f$ is open by assumption). It is easy to see then that $f(U)$ and $V$ are disjoint open neighborhoods of $y_1,y_2$.

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  • $\begingroup$ Hi, I want to clarify how $f(U)$ and $V$ are disjoint open neighborhood of $y_1$, $y_2$. I can see that they are disjoint but I don't know the facts that make them disjoint. Is it because of the chosen distinct points in $Y$? $\endgroup$
    – yaechan
    Jan 14, 2023 at 14:42
  • $\begingroup$ @yaechan It is because $U\times V$ is disjoint from $G_f$. In fact, if $f(u)=v\in f(U)\cap V$ for some $u,v$ then $(u,v)$ is a point of $(U\times V)\cap G_f$, a contradiction. $\endgroup$
    – Ayaka
    Jan 14, 2023 at 14:50
  • $\begingroup$ As above, $f(U)$ and $V$ are disjoint. Since $U$ is an open neighborhood of $x_1$, $f(U)$ is an open neighborhood of $f(x_1)=y_1$ since $f$ is open. (I should have noted that $U$ and $V$ are as such, I'll edit the answer.) $\endgroup$
    – Ayaka
    Jan 14, 2023 at 15:09
  • $\begingroup$ ohh i see, thank u so much! $\endgroup$
    – yaechan
    Jan 14, 2023 at 17:09
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Here is another way to solve it.

Since the inverse image of the diagonal $\Delta \subset Y \times Y$ under the map $f \times \text{id}_Y \colon X \times Y \to Y \times Y$ is the set $G_f$, we have $$ (f \times \text{id}_Y )^{-1}[(Y \times Y) \setminus \Delta] = (X \times Y) \setminus G_f. $$ Now, if $f$ is surjective, then so is $f \times \text{id}_Y$, so that applying $f \times \text{id}_Y$ on both sides of the above equality yields $$ (Y \times Y) \setminus \Delta = (f \times \text{id}_Y)[(X \times Y) \setminus G_f]. $$ Finally, note that if $f$ is open, then so is $f \times \text{id}_Y$, and then the above implies that $\Delta$ is closed in $Y \times Y$, i.e. $Y$ is Hausdorff.

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  • $\begingroup$ +1; The diagonal criterium for Hausdorff spaces always gives you the most elegant proof. $\endgroup$ Jan 15, 2023 at 17:10

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