# Prove that $Y$ is a Hausdorff if $f$ is surjective open map and its graph $G_f$ is closed

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 $$x_2 \in \pi_1(U_2)$$.

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

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 $$y_1,y_2$$.
• 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 $Y$? Commented Jan 14, 2023 at 14:42
• @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. Commented Jan 14, 2023 at 14:50
• 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.) Commented Jan 14, 2023 at 15:09
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.