# The pullback $F^\ast :T^*N \rightarrow T^*M$ is a smooth bundle map

How can I show that the pullback $F^*: T^*N \rightarrow T^*M$ associated with $F:M \rightarrow N$ is a smooth bundle map if it is a diffeomorphism?

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Apply the definition and check? – Ryan Budney Dec 6 '11 at 1:32
I think that you should reverse the arrow in your $F^\ast$, because $F^\ast:\eta_x\in T^\ast M\mapsto \eta_x\circ(T_xF)^{-1}\in T^\ast N\$ is a well defined smooth bundle map over $F$ even when $F$ is just a local diffeomorphism. – Giuseppe Dec 6 '11 at 14:47
I still cannot see the answer. – user20353 Dec 24 '11 at 22:36
Try computing this in local coordinates. If you know what $F^*$ does to $dx^i$ you're done! – Otis Jan 12 '12 at 23:19
Is this a homework problem? – Jon Beardsley Jan 12 '12 at 23:34

Let $q$ be a point in $N$ and $\omega$ be a smooth covector field on N. Assume $F^{-1}(q)=p$. Then $(F^{\ast} w)_p=F^{\ast}(\omega_q)$ is a smooth covector field on $M$. Let's denote the bundle projections by $\pi_M$ and $\pi_N$. Therefore $\pi_M(F^\ast (\omega_q))=p$ and $F^{-1}(\pi_N(\omega_q))=p$. Hence $F^\ast$ covers $F^{-1}$.
Also, since the pullback map $F^\ast$ is dual to the pushforward map $F_\ast$, restriction of the pullback map to each fiber is linear. Moreover, one can show that $F^\ast$ is a smooth map between smooth manifolds. Thus, the pullback $F^\ast$ is a smooth bundle map.
There is a subtlety to the pullback (if the base map is not a diffeomorphism, that is). If $F$ is not surjective, then the domain of $F^*$ will not be all of $T^* N$. Additionally, one must know the preimage of the basepoint in $T^* N$ for the action of $F^*$ to be well defined. Thus $$"F^* \colon T^*N \to T^*M"$$ is ill-defined due to a subtle type error. Considering things fiber-by-fiber, if $p \in M$, then $$(F^*)_p \colon T^{*}_{F(p)}N \to T^{*}_{p}M$$ is just the linear adjoint of $$(F_*)_p \colon T_p M \to T_{F(p)} N.$$
To handle $F^*$ as an "entire" object, the notion of a pullback bundle ( http://en.wikipedia.org/wiki/Pullback_bundle ) must be introduced. This is a way to "pull" the fibers of a bundle back over a different base manifold. In particular it can be used to "remember" the basepoint of the domain element in the pushforward of a map. Apropos to this question, the pullback of $TN$ by $F$ is a vector bundle $F^* TN$ which has base space $M$, and is defined as a submanifold of $M \times TN$ (see the wiki article). Then $$F_* \colon TM \to F^* TN, \, \frac{d}{dt}p(t)\mid_{t=0} \mapsto (p(0),\frac{d}{dt}F\circ p(t)\mid_{t=0}).$$ Here, you can see that the basepoint of the input is encoded in the output. Then, the pullback of $F$ can be constructed naturally as the adjoint of this object. $$F^* \colon F^* T^* N \to T^* M,$$ where the knowledge of the preimage basepoint is encoded in the domain itself.