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?
 A: 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.
A: 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.   
