Let $F:M\to N$ be a smooth map and $\omega$ be a $1$-form on $N$.
$F$ induce a map on the tangent spaces at each point $dF_p:T_pM\to T_{F(p)}N$, and we can take the linear dual of each of these $(dF_p)^*:T^*_{F(p)}N\to T^*_{p}M$.
I want to show that we can define a pull-back differential $1$-form $F^*\omega$ such that $(F^*\omega)_p = F^*(\omega_{F(p)})$.
So I have to show that $F^*\omega:M\to T^*M$ is smooth, and assigns an alternating $1$-form at each point.
To see that this is smooth, I think I can realize it as a composite: $$M\stackrel{F}{\to}N\stackrel{\omega}{\to}T^*N\stackrel{(dF)^*}{\to}T^*M$$ where $F,\omega$ are smooth a priori, so I then just have to show that $(dF)^*$ is smooth to show that $F^*\omega$ is smooth right? But $(dF_p)^*$ just precomposes things, i.e. $(dF)^*(\omega) =\omega\circ dF$ which is a composite of smooth things, so is smooth?
So I can conclude that $F^*\omega$ is smooth?
Does the alternating $1$-form part come vacuuously?