Let $M, N$ be smooth manifolds, $f : M \to N$ a smooth submersion and $g : M \to N$ a smooth function. Is it true that $R = \{x \in M\ |\ f(x) = g(x)\}$ is an embedded submanifold of $M$? How about if both $f, g$ are submersions?
I tried to show that the map $(f, g) : M \to N \times N$ is transverse to the diagonal $\Delta \subset M \times M$, but I can't see it. Am I going crazy here?
It's not true that $R$ is necessarily a manifold, even if $g$ is a submersion. Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ the usual projection map onto the first coordinate, $f(x,y) = x$, which is definitely a submersion.
Suppose $g:\mathbb{R}^2\rightarrow \mathbb{R}$ is defined by $g(x,y) = xy$. Then $R$ consists of all those points with $xy = x$, that is, with $x(y-1) = 0$. This is a union of the $y$-axis together with the horizontal line $y = 1$. In particular, $R$ is not a manifold at the point $(0,1)$.
Now, $g$ is not a submersion because the differential $dg$ at the point $(0,0)$ is not surjective. However, $g$ is a submersion on $\mathbb{R}^2\setminus\{(0,0)\}$. So, if you restrict $f$ and $g$ to $\mathbb{R}^2\setminus \{(0,0)\}$, then $R$ is still not a manifold, but $f$ and $g$ are submersions.