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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

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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
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
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
But I don't think that's true in general. See the example in my answer. – John Klein May 19 '13 at 13:43


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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