Properties of forms and pullbacks Let $j: \mathbb{S}^2 \rightarrow \mathbb{R}^3\setminus\{0\}$ be the canonical injection and $\alpha$ a k-form over $\mathbb{R}^3\setminus\{0\}$.


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*If $\alpha$ is closed or exact, is it the same for $j^*\alpha$?

*If yes, can this be generalized to any regular map: $f: M \rightarrow N$ and any $k$-form over $N$?

 A: The key fact that you need is the naturality of the exterior derivative:

If $F : M \to N$ is a smooth map between smooth manifolds, the pullback map $F^* : \Gamma(\bigwedge^k T^*N) \to \Gamma(\bigwedge^k T^*M)$ commutes with the exterior derivative, that is, for any smooth $k$-form $\alpha \in \Gamma(\bigwedge^k T^*N)$ on $N$, $$\boxed{F^* d\alpha = d(F^* \alpha)} .$$

For more, see, e.g., $\S$ 12 of Lee's Introduction to Smooth Manifolds, where this fact is labeled Lemma 12.16.
A: As @Travis pointed out, you should use the fact that the pullback of differential forms commutes with the differential.
Moreover, if $f: M \rightarrow N$ is a smooth map, then the induced map $f^{*}$ takes closed form to the closed forms and exact forms to the exact forms, which descends to the linear map between cohomology groups of $M$ and $N$
$$f^{*}: H^{n}(M) \rightarrow H^{n}(N)$$
Indeed, let $\omega$ be an exact form, then there exists a form $\eta$ such that $w = d \eta$, then $$f^{*} (\omega) = f^{*} d \eta = d (G^{*} \eta)$$ whilst the latter implies that that the $f^{*}(\omega)$ is also exact.
Now let $\omega$ be a closed form, hence, since the derivative commutes with the pullback
$$d(f^{*} \omega) = f^{*} (d \omega) = f^{*} 0 = 0$$
So, if $[\omega] = [\omega']$, then $f^{*}[\omega] = f^{*}[\omega']$, as desired.
