Let $M$ and $N$ be smooth manifolds ($C^\infty$). Let $s\in C^\infty(M, N)$ and $u\in C^\infty(N, M)$ be maps satisfying:
- $u$ is an embedding;
- $s\circ u=\textrm{id}_N$;
- $(u\circ s)^2=u\circ s$.
Using this data can I conclude $s$ is a submersion?
Thanks.
Not in general. Let $N = \mathbb{R}, M = \mathbb{R}^2$ and consider the following maps:
$$ u(x) = (x,0), \,\,\, s(x,y) = x(y+1). $$
The map $u$ is an embedding whose image is the $x$-axis. The map $s$ is a smooth retract onto the $x$-axis and like Mike Miller noted, the third condition follows from the second condition. Geometrically, $u \circ s$ sends points on the hyperbola $x(y+1) = \varepsilon$ (or $y = \frac{\varepsilon}{x} - 1$) to their unique intersection $(\varepsilon, 0)$ with the $x$-axis. However, the map $s$ is clearly not a smooth submersion since the preimage $s^{-1}(0)$ is a degenerate hyperbola which is not a smooth manifold (or alternatively, $ds|_{(0,-1)} = 0$).
However, since the first condition implies that $ds_p$ must be onto for all $p \in \mathrm{Im}(u)$ and since being onto is an open condition, you can conclude that there exists an open neighborhood $U$ of $\mathrm{Im}(u)$ in $M$ such that $s|_U$ is a submersion.