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.

You need to write $f \times g \in C^{\infty}(X \times X, \mathbb{R}^2)$. Use the definition of smooth, which says that we have to check that given any $(p, q) \in X \times X$, the coordinate representation of $f \times g$ on some coordinate patch $U \ni (p, q)$ is smooth. Pick neighborhoods $U \ni p$ and $V \ni q$ and charts $\phi : O_x \to U$ and $\psi : \Omega_y \to V$, where $O$ and $\Omega$ are open in $\mathbb{R}^n$. then the coordinate representation of $f \times g$ wrt the chart $\phi \times \psi : O \times \Omega \to U \times V$ is $((f \times g) \circ \phi \times \psi)(x, y) = (f(\phi(x)), g(\psi(y)))$. Now by assumption that $f, g$ are smooth, $f^x = f \circ \phi : O \to \mathbb{R}$ and $g^y = g \circ \psi : \Omega \to \mathbb{R}$ are smooth. By the formula $$\frac{\partial}{\partial x_j} (f^x(x), g^y(y)) = (\partial_{x_j}f^x(x), g^y(y))$$
and other analogous formulas and the fact that a $\mathbb{R}^k$ valued function is smooth if and only if all it's components are smooth, we have $f^x \times g^y \in C^{\infty}(O \times \Omega, \mathbb{R}^2)$ as desired.
