Problem: A fixed point of a function $f: X \to X$ is a point $x_0 \in X$ such that $f(x_0) = x_0$. The diagonal of $X$ is defined as
\[
\Delta (X) = \left\{ (x,x): x\in X \right\} \subseteq X \times X.
\]
Prove that $f$ has a fixed point if and only if graph of $f$ intersects the diagonal of $X$ non-trivially.
Solution: Let $\mathop{\mathrm{Gr}}(f) $ denotes the graph of $f$ which is defined as
\[
\mathop{\mathrm{Gr}}(f) \coloneqq \left\{ (x,f(x)) : x\in X \right\} .
\]
Let $f$ has a fixed point $x_0$. That is, $f(x_0) = x_0$. Thus,
\[
\Delta (X) \ni (x_0, x_0) = (x_0, f(x_0)) \in \mathop{\mathrm{Gr}}(f) .
\]
On the other hand, let $\mathop{\mathrm{Gr}}(f) \cap \Delta (X) \neq \emptyset $. Let
\begin{align*}
(x,y) \in \mathop{\mathrm{Gr}}(f) \cap \Delta (X) & \implies (x,y) \in \mathop{\mathrm{Gr}}(f) \text{ and } (x,y) \in \Delta (X) \\
& \implies y = f(x) \text{ and } y = x \\
& \implies f(x) = x.
\end{align*}
Therefore, $x$ is a fixed point of $f$.