Let $M$ be a smooth manifold and $\Delta_M=\{(p,p)\,\vert\, p\in M\}$ be the diagonal of $M$. Then $\Delta_M$ is a submanifold of $M\times M$.
I have proved this by showing that $\mathscr{i}:M\to M\times M,\, \mathscr{i}(p)=(p,p)$ is an embedding. Otherwise it also shouldn't be difficult to show that it has the submanifold property (i.e. we have adapted coordinates at every point). However, I came across a different argument:
For $(\varphi,V)$ chart around $p\in M$ consider $f:V\times V \to \mathbb{R}^n,\,f(p,p')=\varphi(p)-\varphi(p')$ which is a submersion. Locally we then have $\Delta_M=f^{-1}(0)$ and consequently it's a submanifold of $M\times M$.
I know for a smooth map between manifolds $f:N\to M$ with constant rank on $N$ that $f^{-1}(q)$ is a closed submanifold of $M$, but I haven't figured out how this above condition is compatible. [I suspect that the argument isn't too complicated]
This is quite reminiscent of the definition of submanifolds in $\mathbb{R}^n$ and that's what leads me to believe that we have this local zero set description for submanifolds in a general way. So my question is actually more general.