Proving the image of a surjective function is Hausdorff

Question

Show that if $f:X\rightarrow Y$ is an open surjection such that $Gf$, graph of $f$, is closed, then $Y$ is Hausdorff.

What does it mean $Gf$ closed?

Answer 1

Set-theoretically, a function $f\colon X\to Y$ is (along with knowledge of the domain and codomain) a subset of $X\times Y$, call the graph $Gf=\{(x,f(x))|x\in X\}.$

When dealing with topological spaces, now $Gf$ is a subset of the topological space $X\times Y$, endowed with the product topology. As such we may ask whether it may be open or closed or both or neither, as in any topological space. In this context, it means what it always means; that the limit of every set in $Gf$ is in $Gf$. That the complement is open, so that every point in $X\times Y\setminus Gf$ has an neighborhood entirely within $X\times Y\setminus Gf$.