Let $f:N\rightarrow M$ be a surjective submersion.
Let $g:W\rightarrow M$ be a smooth map.
Let $w\in W$. We have $g(w)\in M$. As $f$ is surjective, there exists $n\in N$ such that $f(n)=g(w)$.
Define $p:W\rightarrow N$ such that $p(w)=n$ where $n\in N$ is such that $f(n)=g(w)$.
I am wondering if this map $p$ is smooth.
Any hints are welcome.
I am seeing the situation as following case
As $f$ is a submersion, pull back $W\times_MN$ is a smooth manifold.
As $f$ is surjective, given $w\in W$ there exists $n\in N$ : $f(n)=g(w)$, giving an element $(w,n)\in W\times _MN$ which gives a obvious map $W\rightarrow W\times_M N$ and it is clear that $pr_2$ composed with this map is equal to $p$.
As $pr_2$ map is smooth and as $W\rightarrow W\times_MN$ is smooth (being an inclusion) the composition $p:W\rightarrow N$ is smooth.
Flaw in my idea is that the map $W\rightarrow W\times_MN$ is not an inclusion. It is not of the form $w\mapsto (w,n)$ for some fixed $n$. It is of the form $w\mapsto (w,n_w)$ which is not inclusion.
Any suggestions to make this work are welcome.