Corollary 11.46 of John Lee's introduction to smooth manifold. The original satatement is that

Suppose that $F : M → N$ is a $\textbf{local diffeomorphism}$. Then the pullback $F^∗:\mathfrak{X}^*(N)\to \mathfrak{X}^*(M)$ takes closed covector fields to closed covector fields, and exact ones to exact ones.

The proof of $F^*$ sends exact one to exact one does not use the condition $F$ is a local diffeomorphism. It's used in the proof for the case of closed covector fields.
And now my professor said that we can drop the local diffeomorphism condition if $M := U\subset\Bbb R^n$ and $N:=V\subset\Bbb R^m$. I don't understand why is true in that case. Could anyone explain this?
 A: If $U \subset \mathbb{R}^n$ and $V\subset \mathbb{R}^m$ are open subsets and if $F : U \to V$ is smooth, one can define
$$
F^* : \Omega^1TV \to \Omega^1TU
$$
by $(F^*\alpha) = \alpha\circ F_*$, where $F_* = \mathrm{d}F$ is the differential of $F$. It does not require any condition on the rank of $F_*$ to be defined. Moreover, the same construction can be applied to construct $F^* : \Omega^p TV \to \Omega^pTU $ for all integers $p \geqslant 0$. One can show that $F^* : \Omega^*TV \to \Omega^* TU$ is thus well defined and commutes with the exterior differential (this last claim is basically the chain-rule). We often say that $F^*$ is "natural". Hence, if $\alpha$ is a 1-form (a covector):
$$
\mathrm{d}\left( F^*\alpha\right) = F^* \left( \mathrm{d}\alpha\right),
$$
which shows that if $\alpha$ is closed (that is $\mathrm{d}\alpha = 0$), then so if $F^*\alpha$ (i.e $\mathrm{d}\left(F^*\alpha\right)=0$).
Note that the same calculation show that if $\alpha$ is an exact 1-form, say $\alpha = \mathrm{d}f$, then:
$$
F^*\alpha = F^*\left(\mathrm{d}f \right) = \mathrm{d}\left(F^*f \right) = \mathrm{d}\left(f\circ F\right),
$$
and $F^*\alpha$ is exact.
Comment Notice this construction is local, hence is still true for $F : M \to N$ a smooth map between smooth manifolds.
