Critical Points of a smooth map on a compact manifold Show that a smooth map $f$ from a compact manifold $N$ to $\mathbb R^m$ has a critical point. (Hint: Let $\pi$: $\mathbb R^m \rightarrow \mathbb R$ be the projection to the first factor. Consider the composite map $\pi $ $\circ$ $f$: $N \rightarrow \mathbb R$.
Assumptions given:
$N$ is a compact manifold
$\pi $ $\circ$ $f$: $N \rightarrow \mathbb R$.
$\pi$: $\mathbb R^m \rightarrow \mathbb R$
$f$ is a smooth map.
Attempt at proof:
Let $p \in N$. Smoothness of $f: N \rightarrow \mathbb R^m$ means that there are charts $(U,\phi)$ about $p \in N$ and $(V,\psi)$ about $f(p)$ $\in \mathbb R^m$ such that $\psi \circ f \circ \phi^{-1}$ is smooth. 
I was pondering if I should use a chart $(V,\pi)$ in $\mathbb R^m $ and possibly $(U, \pi \circ f)$ in $N$. I'm also wondering how to use the compactness of N. I'm a bit lost, so without giving the full solution, I'm looking for a push in the right direction via some extra advise on how to approach this problem.  
BTW I'm using Introduction to Manifolds by Loring Tu.
 A: The hint given to you is related to the fact that every smooth real function (i.e., $f:M \to \mathbb{R}$) on a compact manifold has a critical point. This is, in turn, due to the fact that every continuous real function on a compact set attains a maximum at some point $p$. Then, if the function is differentiable on this point $p$, we must have that the derivative is zero (if you don't know how to prove this, you can pick curves passing through $p$ and use the one-variable result of "local maximum$\implies f'=0$". The next step depends on your explicit definition of derivative on manifolds).
Given that, one now uses the chain rule in your case. Indeed, letting $p$ be a critical point of $\pi \circ f$, we know that $(\pi \circ f)'_p=\pi'_{f(p)} \circ f'_p=0,$ which implies that $f'_p$ cannot be surjective, since composition of surjective linear maps $(\pi'_{f(p)}$ is obviously surjective) is surjective.
Just to be extremely explicit, since this may cause confusion: a critical point is not a point where the derivative is zero, in general. It is a point where the derivative is not surjective. These concepts coincide in the case of codomain with dimension $1$ because of the fact that linear maps with $1$-dimensional codomain are either surjective or zero.
