If $\pi\colon X\to Y$ is a quotient map, and $f\colon Y\to Z$ is a continuous map between topological spaces, then the characteristic property of the quotient map says $f$ is continuous iff $f\circ \pi$ is continuous.
Does this also work for smooth maps if the spaces in question are all smooth manifolds? Would $f\circ \pi$ smooth imply $f$ is smooth too? The characteristic property for continuous maps is well-documented, but I couldn't find anything for smooth maps.
The analog of a quotient map in the smooth category is a surjective smooth submersion. It satisfies the following characteristic property:
Suppose $X$, $Y$, and $Z$ are smooth manifolds and $\pi\colon X\to Y$ is a surjective smooth submersion. Then a map $f\colon Y\to Z$ is smooth if and only if $f\circ\pi$ is smooth.
This is Theorem 4.29 in my Introduction to Smooth Manifolds (2nd ed.).
Here is a counterexample of sorts. Take the circle and identify the top and bottom arc to obtain $[-1,1]$. Near $\pm 1$, the folding maps locally look like the function $x^2$. Now identify the endpoints $\pm 1$ to get a topological quotient map from $S^1 \rightarrow S^1$. Let $P$ be the point resulting from the endpoints being glued. In suitably chosen local coordinates around the critical value $P$, the quotient map looks like $-(x-2)^2, (x+2)^2$ being projected onto the $y$-axis. In this same chart, consider the map $|x|$ on the quotient circle, suitably extended so that it's smooth everywhere but $P$. This map is not smooth at $P$, but the composition with the quadratic functions is smooth.
So we see that being a submersion is essential in the statement that Jack Lee posted.
