# 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