Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 3 / 3G (Further Analysis)

Take $(x_0,y_0)$ outside the graph of a function, so $y_0\ne f(x_0)$. Since $Y$ is a Hausdorff space, choose disjoint open neighborhoods $V$ of $y_0$ and $W$ of $f(x_0)$. Continuity makes $U=f^{-1}(W)$ an open neighborhood of $x_0$. The product neighborhood $U\times V$ contains $(x_0,y_0)$ and cannot meet the graph, because every $f(x)$ with $x\in U$ belongs to $W$. The complement is therefore open in the product topology, provingThis is the closed graph of a map into a Hausdorff space property; no separation assumption on $X$ is required.

The map $h:X\to X\times Y$, $h(x)=(x,f(x))$, is continuous: the inverse image of a basic open set $U\times V$ is $U\cap f^{-1}(V)$. Its image is exactly the graph of a function $G_f$. The continuous image of a compact space is compact, soThe compact graph of a continuous map conclusion does not need $Y$ to be Hausdorff. In fact projection onto $X$ is a continuous inverse of $h:X\to G_f$, so the graph is homeomorphic to $X$.