Question about rank of a smooth map. Suppose $M$ is a compact smooth manifold without boundary of dimension $n$, and given a smooth map $f:M\to \mathbb{R}^n$. Prove that there at least one point $p$ on $M$, s.t. the rank of $f$ on $p$ is smaller than $n$.
Here is my attempt. If all point on $M$ has full rank, then $f$ is a map with constant rank $n$. By rank theorem, every point on $M$ is the image of a section $s_i$. Since $M$ is compact, it's closed, and $f(M)$ is closed by rank theorem. We can choose finite sections {$s_i$}, whose images cover $M$. And the closure of the union of their domains is a manifold with boundary in $\mathbb{R}^n$, which is also $f(M)$. But this is impossible because $M$ has no boundary.
I am not sure about my argument. Hope someone could help. Thanks!
 A: I'm not sure about your argument, at least in this formulation. Of course, the boundaryless assumption is crucial in this statement, but I would rather use a direct argument instead of a contradiction agument.
Recall that if $g : M \to \mathbb{R}$ is continuous and if $M$ is compact, then it has a global maximum. If $M$ has no boundary and if $g$ is smooth, then at this point, say $p$, is a critical point, i.e $\mathrm{d}g(p) = 0$.
Write the function $f$ as
$
f = \left(f_1,\ldots,f_n\right)
$ where $f_i : M \to \mathbb{R}$ is smooth. Then in this way of writing things
$$
\mathrm{d}f = \left(\mathrm{d}f_1,\ldots,\mathrm{d}f_n \right),
$$
and the rank of $\mathrm{d}f(x)$ at a point $x$ is at most the sum of the ranks of each $\mathrm{d}f_i(x)$.
As $M$ is compact and $f_1$ is continuous, it achieves its maximum at a point $x \in M$. As $M$ is without boundary, at such $x \in M$, $\mathrm{d}f_1(x) : T_xM \to \mathbb{R}$ is the zero linear function ($x$ is a critical point of $f_1$). Thus, at $x$, the rank of $\mathrm{d}f(x)$ is at most $n-1$.
Edit Here is a proof by contradiction that looks a -very little- bit like yours. Suppose $f : M^n \to \mathbb{R}^n$ is smooth, with $M$ compact without boundary. Suppose by contradiction that for every $x \in M$, the rank of $\mathrm{d}f(x)$ is $n$. By the inverse function theorem, $f(M)$ is open in $\mathbb{R}^n$. But as $M$ is compact, $f(M)$ is compact, hence closed in $\mathbb{R}^n$. Thus, $f(M)$ is open and closed, an non-empty: by connectedness, $f(M) = \mathbb{R}^n$. But $M$ is compact, so is $f(M)$, which leads to a contradiction.
A: Your argument looks roughly ok, but there are simplifications that can be made that will make this cleaner. First of all, $f(M)$ is compact simply because $M$ is. Choosing a point of $f(M)$ of maximal distance from the origin, it must be a boundary point. At this point, $f$ fails to be a local homeomorphism, and so it will fail to have full rank somewhere (for the reasons you already stated).
