Show that there may not exist a vector field $Y$ on $N$ such that $f_\ast(X_p) = Y_{f(p)}$.

Question

I'm trying to come up with some simple examples where, given a function $f: M \to N$ and a vector field $X$ on $M$, there may not exist a vector field $Y$ on $N$ such that $f_\ast(X_p) = Y_{f(p)}$.

I know the sufficient condition for $f$-relatedness of vector fields is that $f$ be a diffeomorphism, so I should be able to pick a function $f$ which is not a diffeomorphism and any vector field $X$ (perhaps a coordinate vector field) for my example, but I'm having trouble seeing all the details.

Answer 1

Simplest example:

Let $M = (0,1) \cup (2, 3)$, $N = (0,1)$ and $N = (0,1)$. Let $f:M\to N$

$$ f(x) = \begin{cases} x & \text{if } x\in (0,1),\\ x-2 & \text{if } x\in (2, 3).\end{cases}$$

Let $X$ be nonzero $(0,1)$ and $0$ on $(2, 3)$. Then $f_*(X_x)$ is non-zero on $x\in (0,1)$ and is zero on $(2, 3)$. Then

$$ f_*(X_{0.5}) \neq f_*(X_{2.5})$$

and thus there isn't such a vector fields $Y$.