Consider the following definition:
If $U \subset \Bbb{R}^{m+n}$ and $V \subset \Bbb{R}^m$ are open sets, $\xi: U \to \Bbb{R}^{m+n}$ is a vector field and $f: U \to V$ a surjective submersion, the push-forward of $\xi$ by $f$ (if it exists) is the only vector field $f_{\ast}\xi: V \to \Bbb{R}^m$ such that, for each $x \in U$, $$(f_{\ast}\xi)(y) = Df(x)\xi(x)$$ where $y = f(x)$.
I undestood the definition, but I could not find some submersion that has no push-forward. I would like to know some example and why the push-forward does not exist. Thanks for the advance!
The vector field is defined on $V$, so it should have a well-defined value in each $y\in V$. However, that value depends on a pre-image $x$, so if that is not unique, the vector field is only well-defined when the derivative of $f$ is the independent of the specific element in the pre-image $f^{-1}(\{y\})$ of each point $y$.
Possibly the easiest example where it is not well-defined would be $U = (0,1/2) \cup (1,2)$, $V = (0,1)$, defined by $f(x) = 2x$ for $0 < x < 1/2$, and $f(x) = x - 1$ for $1 < x < 2$, and $\xi \equiv 1$.