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 $U$.



Vector field pull back from embedding
At each point $x\in M$ the differential $df\_x: T\_x M \to T\_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X\_{f(x)}$ need not be in the image of $df\_x$. Hence to associate a tangent vector to $M$ at $x$, you need a procedure which associates to a vector in the vector space $T\_{f(x)}N$ a vector in the subspace $df\_x(T\_xM)$. To have the correct properties, this vector should coincide with the given vector in $T\_{f(x)}N$ when $\dim M = \dim N$. How is that to be accomplished?


For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field with value $\langle 1,-1 \rangle$, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).


One way to do this, which is not canonical, is
to choose a splitting $f^\*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project
$X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).


**Added Later:**


I didn't read the question as carefully as I should have.


The submitter's choice of $r: U \to M$ amounts to 
the choice of a smooth retraction
of a tubular neighborhod of $f(M)$ to $f(M)$. 


The space of such choices is contractble, but I doubt that there is
a preferred basepoint in this space of choices.


Furthermore, the vector field you get on $M$ depends crucially on the choice of $r$.


**Example:** in the $\Bbb R^2$ example, let's take $U = \Bbb R^2$ and defined
$r: U \to \Bbb R$ to be (i) the first factor projection, or (ii) the map $(x,y) \mapsto x-y$. If $X$ is the vector field which is $\langle 1,1\rangle$ at every point of $\Bbb R^2$, then $r\_\*$ applied to $X$ gives the constant unit vector field on the $x$-axis in case (i) and the trivial vector field in case (ii). 


(By the way, the retraction induces a splitting $f^\*TN \cong TM \oplus \nu$ of the kind
mentioned above.)


