About the continuity of a function in the closed graph theorem proof

Question

I'm reading Functional Analysis book of Rudin, and in the proof of the closed graph theorem, there's one point that I don't understand. Can someone please explain it to me? I really appreciate this. Thanks

$X ,Y$ are $F$-spaces, $f: X \rightarrow Y$ is linear, $G = \{(x, f(x)): x \in X\}$ is closed in $X \times Y$. Then the mapping $\pi:G \rightarrow X$ defined by $\pi(x, f(x)) = x$ is continuous.

The author states as though it's obvious, but I don't know why. Please don't use the result of the closed graph theorem, because this question is in the proof of it. Thanks

Answer 1

For any two topological spaces $X$ and $Y$, the projections

$$\pi_X \colon X\times Y \to X;\quad \pi_Y \colon X\times Y \to Y$$

are continuous when $X\times Y$ is endowed with the product topology.

Thus the restrictions of the projections to any subspace of $X\times Y$ are continuous too.

Here, $\pi$ is the restriction of $\pi_X$ to the graph $G$.