Let $M$ and $N$ be finite dimensional smooth manifolds.

A smooth map $f: M \to N$ is an embedding if and only if there is an open neighborhood $U$ of $f(M)$ in $N$ and a smooth mapping $r : U \to M$ with $r \circ f = Id_M$.

Does this mean we can pull back a vector-field $X$ on $N$ to a vector field on $M$, like we could, if $f$ were a diffeomorphism?

It seems like we can define the vector field $Y$ on $M$ by

$$ Y(m):=r_*(X(f(m))) $$

Any problems with that? (I'm just wondering because until now I though that we can use only diffeomorphisms to pull back vector fields, but it seems that this weaker condition is in fact enough. Or what am I overlooking?)

Let $M$ and $N$ be finite dimensional smooth manifolds.

A smooth map $f: M \to N$ is an embedding if and only if there is an open neighborhood $U$ of $f(M)$ in $N$ and a smooth mapping $r : U \to M$ with $r \circ f = Id_M$.

Does this mean we can pull back a vector-field $X$ on $N$ to a vector field on $M$, like we could, if $f$ were a diffeomorphism?

It seems like we can define the vector field $Y$ on $M$ by

$$ Y(m):=r_*(X(f(m))) $$

Any problems with that? (I'm just wondering because until now I though that we can use only diffeomorphisms to pull back vector fields, but it seems that this weaker condition is in fact enough. Or what am I overlooking?)

EDIT: An appropriate negative answer has to clarify, why the particular choice of $r$ matters here. From $r\circ f = id_M$ we get that on $f(M)$ $r'=r$ for any two such maps and hence $r_{*|f(M)}=r'_{*|f(M)}$. So the only thing that really can be non natural here could be some wired behavior on the boundary between $f(M)$ and $U$.