If the restriction of a differential form to a closed submanifold is closed, is the form itself closed?

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Let $X$ be a smooth manifold and let $i:Y\hookrightarrow X$ be a closed embedding of a closed submanifold $Y$ in $X$. It is well-known that the sheaves of smooth differential operators $\Omega_X^{p}$ are fine, which means that they are soft, in particular. That is, the map

$$i^{*}:\Omega^{p}_X(X)\to i_*i^{*}\Omega_X^p(X)=\Omega_{X|Y}^p(Y)$$

is surjective. Assume now that $\alpha$ is a closed $p$-form on $Y$. Then, by the softness property, there is a $p$-form $\beta$ on $X$ such that $i^*\beta=\alpha$. Since $\alpha$ is a cocycle, we have

$$[\alpha]=[i^*\beta].$$

Is it true that

$$[\alpha]=i^*[\beta]?$$

We know that the pullback commutes with the differential but $\beta$ is just a p-form, it is not closed, so the class $[\beta]$ of an arbitrary form $\beta$ does not make sense, unless $\beta$ is a closed form, which is not guaranteed.