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 $x_2 \in \pi_1(U_2)$.
Best Answer
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 $y_1,y_2$.