# Pulling back vector fields

I want to find conditions under which one can pull-back vector fields (if it is at all possible).

Let $F:M \to N$ be a smooth surjective map between two $C^{\infty}$ manifolds of the same dimension. Let $Y$ be a vector field on $N$ (i.e. smooth section of the tangent bundle $TN$). Define: $T^\ast Y(p)(f)=Y(F(p))(f\circ F^{-1})$, where $f \in C^{\infty}(M,\mathbb R)$. We check that $T^\ast Y$ is a derivation, and this is true since:

$T^\ast Y(p)(fg)=Y(F(p)(fg \circ F^{-1})=Y(F(p)((f \circ F^{-1})(g \circ F^{-1}))$

My question: Is it enough for $F$ to be a local diffeomorphism for this to work?

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How is $f \circ F^{-1}$ an element of $C^\infty(N,\mathbb{R})$, if you don't assume that $F^{-1}$ is a smooth map $N \to M$? In other words, it appears to me that you need to assume that $F$ is a diffeomorphism, and then, AFAICT, your definition reduces to the pushforward of $Y$ under $F^{-1}$. – Martin Wanvik Apr 27 '12 at 16:59
@MartinWanvik: Actually, I just need $f \in C_p^{\infty}(M,\mathbb R)$, i.e, the germ at $p$. I am trying to avoid a full diffeomorphism. – Confused1001 Apr 27 '12 at 17:32
Right, of course - it's been a long time since I've studied this stuff. It might make things clearer if you put some square brackets around the functions, to indicate that you're considering germs instead of actual functions, i.e. $[f \circ F^{-1}]_p$. As you probably know, there are several definitions of tangent spaces, and the one I looked at says that a tangent vector is a derivation on $C^\infty(M,\mathbb{R})$ - of course, it comes out to the same thing, but in that context your formula would require a word or two of explanation. – Martin Wanvik Apr 27 '12 at 17:59
So, from what I can tell, there appears to be no question that question that you're able to associate a derivation to each point in $M$, so what remains is to prove smoothness. As an alternative, if you assume (for example), that both manifolds are Riemannian, you can pull back vector fields without any conditions on the map $F$ (only smoothness). – Martin Wanvik Apr 27 '12 at 18:30
@MartinWanvik: Unfortunately that wont work for my purposes, one of the manifolds is $\mathbb R P^n$ – Confused1001 Apr 27 '12 at 18:37

To pull back vector fields on $N$ you'll want for $F : M \to N$ to be a local diffeomorphism (at each point of $M$):
If $Y$ is a smooth vector field on $\operatorname{Im} F$ (i.e. the image of $F$, which is an open subset of $N$) then you can define a vector field on $M$ by $m \in M \mapsto \left( \operatorname{T}_m F \right)^{-1}(Y_{F(m)})$, which is well-defined since each $\operatorname{T}_m F : \operatorname{T}_m M \to \operatorname{T}_{F(m)} N$ is a bijection. $X$ will be smooth since for any $m \in M$ we can select an open (in $M$) neighbourhood $U$ of $m$ such that $F\big\vert_U : U \to F(U)$ is a diffeomorphism onto the open subset $F(U)$ of $N$. Note that $X\big\vert_U$ is equal to the pushforward $X\big\vert_U = \left( F\big\vert_U^{-1} \right)_{*}\left( Y\big\vert_{F(U)} \right)$, which is smooth so that $X$ is smooth around each point of $M$.
Edit: Using the above result, by going into coordinates it is then straightforward to show that if $F : M \to N$ is an immersion and $Y_{F(m)} \in \operatorname{Im} \left( \operatorname{T}_m F \right)$ for all $m \in M$ then the map $X : M \to \operatorname{T} M$ defined by $m \mapsto \left( \operatorname{T}_m F \right)^{-1}(Y_{F(m)})$ is a well-defined smooth vector field on $M$ that can be pushed-forward to $Y\big\vert_{\operatorname{Im} F}$. Note that if $\operatorname{Im} F$ is dense in $N$ then the continuous map $Y$ is completely determined by $Y\big\vert_{\operatorname{Im} F}$ so that in this sense, $X$ could be viewed as "essentially" being the pullback of the entire vector field $Y$.