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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 $U$.

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closed as off topic by Benoît Kloeckner, John Klein, Misha, Lee Mosher, Andreas Blass May 20 '13 at 17:31

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This is basic differential geometry, not research-level. I thought that Lev Soukhanov answer would show you where is the problem, but now the ongoing discussion does not belong here. Voting to close. –  Benoît Kloeckner May 19 '13 at 14:31

2 Answers 2

up vote 3 down vote accepted

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.)

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Why has $X_{f(x)}$ to be in the image of $df_x$? –  Nevermind May 19 '13 at 12:29
    
Not a good style to first downtalk like "by the way ... makes no sense" and then not even say, WHY it makes no sense. –  Nevermind May 19 '13 at 12:36
    
@Nevermind: Concerning your second comment: you're right. I've changed that. Concerning your first one: I meant without making additional choices. Of course, one wishes the associated vector field on $M$ to have certain properties. I could have stupidly defined the "pullback" to $M$ by taking the trivial vector field, but that is not the correct thing to do in the equi-dimensional case. –  John Klein May 19 '13 at 12:47
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Since $r\circ f=id_M$ every choice of $r$ should give the same $r_*$ on $f(m)$. And in the equi-dimensional case this is just the ordinary pullback of a vector field along a diffeomorphism, since embeddings are diffeomorphisms then... –  Nevermind May 19 '13 at 12:52
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But I don't think that's true in general. See the example in my answer. –  John Klein May 19 '13 at 13:43

Edit

Actually, you are projecting your vector field along $r$. This is noncanonical ($TM$ is embedded in $f^*TN$, but $r$, additionally, gives you a splitting, and it is noncanonical part).

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In other word, your "pull-back" vector field depends on both $f$ and $r$, while to properly define a $f^*X$ you would like it to depend only on $f$. –  Benoît Kloeckner May 19 '13 at 11:46
    
Ok it requires both the data of $f$ and $r$ and $r$ is not uniquely defined by $f$. But it is a well defined vector field on $M$ at least. Right? –  Nevermind May 19 '13 at 12:32
    
So what you say is,that different choices of $r$ gives different vector fields on $M$? Hmm... Can you explain why? –  Nevermind May 19 '13 at 12:44

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