Let $f: X \to Y$; let $Y$ be compact Hausdorff. Then $f$ is continuous if and only if $G_f = \{(x,f(x)) \mid x \in X \}$ is closed.
Here is my shot at a proof:
Suppose that $f : X \to Y$ is continuous, and let $(x,y) \in \overline{G_f}$ but assume $y \neq f(x)$. As $Y$ is Hausdorff, there exists $U,V$ open in $Y$ and disjoint such that $y \in U$ and $f(x) \in V$. Since $V$ is a nbhd of $f(x)$ and $f$ is continuous, there exists an $O \subseteq X$ that is open and contains $x$ such that $f(O) \subseteq V$. Since $O \times U$ is an open nbhd of $(x,y)$, there must be some $(p,q) \in G_f \cap O \times U$. This means $p \in O$ and that $f(p)=q \in U$. But $f(O) \subseteq V$, so $U$ and $V$ must intersect–a contradiction. Hence $y= f(x)$ and $G_f$ must be closed.
Now we show the other direction. Suppose that $G_f$ is closed, let $x \in X$ be arbitrary, and let $V$ be a open nhbd of $f(x)$. Then $Y-V$ is closed, and therefore the intersection $C:=G_f \cap [X \times (Y-V)]$ is closed in $X \times Y$. By an earlier problem, we know that $\pi_1$ is a closed map and so $\pi_1(C)$ is closed in $X$, and therefore $X-\pi_1(C)$ is open. I will argue that $x \in X-\pi_1(C)$ by contradiction. If $x$ were in $\pi_1(C)$, then there would exist a $p \in X$ and a $q \in Y-V$ such that $f(p)=q$ and $x= \pi_1(p,f(p))$ or $x=p$. But this would mean $f(x)=q \in Y-V$ or $f(x) \notin V$, which contradicts the assumption that $V$ is a nbhd of $f(x)$. Hence $x \in X – \pi_1(C)$.
Now we argue that $a \in f^{-1}(V)$ if and only if $a \notin a \notin \pi_1(C)$. Suppose that $a \in \pi_1(C)$. Then $a = \pi(p,f(p))$ or $a=p$ where $f(p) \in Y-V$. This means that $f(a) \in Y – V$ and therefore $a \notin f^{-1}(V)$. Now suppose that $a \notin f^{-1}(V)$. Then $f(a) \notin V$ and therefore $f(a) \in Y-V$. From this we get $(a,f(a)) \in G_f \cap [X \times (Y-V)] = C$ which implies $a \in \pi_1(C)$.
From this we can conclude that if $a \in X-\pi_1(C)$, then $f(a) \in V$, proving that $f(X-\pi_1(C)) \subseteq V$ which in turn proves that $f$ is continuous.
How does this sound?
Best Answer
If you know that $\pi_X: X \times Y \to X$ is a closed map (which you seem to):
Suppose $G_f$ is closed.
Let $C \subseteq Y$ be closed. Then $G_f \cap (X \times C)$ is closed in $X \times Y$ and note that $\pi_X[G_f \cap (X \times C)] = f^{-1}[C]$ so that $f^{-1}[C]$ is closed in $X$, as $\pi_X$ is closed. So $f$ is continuous. (inverse image of closed is closed). This direction only uses compactness of $Y$.
For the other direction we only need the Hausdorffness of $Y$: The diagonal $\Delta_Y = \{(y,y) : y \in Y\} \subseteq Y \times Y$ is closed iff $Y$ is Hausdorff, and $G_f = (f \times 1_Y)^{-1}[\Delta_Y]$, where $1_Y$ is the identity on $Y$ and $f \times 1_Y : X \times Y \to Y \times Y$ defined by $(f \times 1_Y)(x,y) = (f(x), y)$ is continuous whenever $f$ is.