Smooth self maps of compact manifolds. Suppose $M$ is a compact $n$ dimensional manifold. Does there exist an example of the following:
A smooth map $f: M \rightarrow M$ such that there exists $x \in M$ where $\mbox{d}_x f$ has maximal rank but $f$ is not surjective.
If there are no examples, how do we prove it? If such examples do exist, is it true that every $f: M \rightarrow M$ with a point of maximal rank is isotopic to a surjective map? What about a diffeomorphism?
 A: Project $S^2$ to the plane, then compose with an embedding $\Bbb R^2 \to S^2$. This is a submersion on each of the hemispheres. You can do a similar construction for any smooth manifold $M$; pick a smooth map $M \to \Bbb R^n$ with surjective differential at some point. (That we can do this is an exercise.) Now compose with a smooth embedding $\Bbb R^n \hookrightarrow M$.
Every smooth map $M \to M$ is homotopic (what does isotopic mean here?) to a surjective map, because there is a surjective smooth map $\Bbb D^n \to M$ for any closed smooth $n$-manifold $M$. So homotope your map inside some ball.
The examples described above are not diffeomorphisms or homotopic to diffeomorphisms, because they induce zero on the top homology group $H_n(M;\Bbb Z/2)$. This is unsurprising. You can't say anything about the map itself just from "$df_x$ is an isomorphism somewhere".
A: Of course there is. Compose the restriction of the projection $(x,y,z)\in\mathbb R^3\mapsto (x,y)\in\mathbb R^2$ to the sphere $S^2$ with the map $(x,y)\mapsto(x/2,y/2)\in\mathbb R^2$ with the map $(x,y)\in(x,y,\sqrt{1-x^2-y^2})\in S^2$.
