Cartesian product of smooth maps is smooth Suppose $f,g \in C^\infty(X)$.  I would like to show that $f \times g \in C^\infty(X \times X)$. This seems obvious, but I'm not sure how to actually prove it. I was thinking of using projection maps $\pi_1$ and $\pi_2$, but I am not sure if this is the right approach.

By definition of $f\times g$ we get that
$f\times g=\alpha \circ (f,g)$
where $\alpha \colon \mathbb{R}^2\to \mathbb{R}$ sends $(x,y)\to xy$ and $(f,g)\colon X\times X\to \mathbb{R}^2$ sends $(p,q)$ to $(f(p), g(q))$.
Of course $\alpha $ is smooth.
Regarding $(f,g)$, the map is smooth if and only if its factors $\pi_i\circ (f,g)\colon X\times X\to \mathbb{R}$ are smooth, where $\pi_i\colon \mathbb{R}^2\to \mathbb{R}$ is the projection map.
Moreover you can prove that $\pi_1\circ (f,g) $ is smooth if and only if $f$ is smooth (the same holds for $\pi_2\circ (f,g)$ and $g$).
Hence, under you assumption of smoothness of $f$ and $g$, then $f\times g$ is smooth.
