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Let $(M,F)$ be a manifold with integrable subbundle $F$ of the tangent bundle $TM$.(foliation).

Q For a submersion $N\to M$, we can lift a subbundle $F'$ of $TN$, can we say $F'$ is still integrable, just because the Bott connection which implies that $TM/F$ is leafwise flat, i.e. $\nabla^{TM/F,2}_{Bott}\mid_\mathcal F=0$.

I do not know the reason.

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    $\begingroup$ Pull back closed differential forms on $N$ that vanish precisely on the leaves of the foliation. Then they are closed differential forms on $M$ that vanish precisely on the leaves of the foliation. $\endgroup$ – Ben McKay Jan 11 '18 at 10:02
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By the Frobenius theorem, any smooth foliation $F$ of any $n$-dimensional manifold $M$, say of rank $k$, has tangent bundle a smooth rank $k$ bracket-closed vector subbundle $TF \subset TM$, and conversely every smooth rank $k$ bracket-closed vector subbundle $V \subset TM$ is the tangent bundle $V=TF$ of a unique smooth foliation $F$. The set of 1-forms $\xi$ on $M$ vanishing on all tangent vectors in $TF$ is a rank $n-k$ subbundle $T^{\perp} F \subset T^*M$, as we can see clearly in any chart in which the foliation is a product. Conversely, a smooth rank $n-k$ vector subbundle $W \subset T^*M$ has the form $W=V^{\perp}$ for a unique rank $k$ subbundle $V \subset TM$. Moreover, $V$ is bracket closed just when, for any local section $\xi$ of $W$, $d\xi$ is a multiple of local sections of $W$. In any coordinates in which a smooth submersion $f \colon N \to M$ becomes a linear submersion, the pullback operation $f^*$ takes $d$ to $d$. Moreover, any constant rank subbundle $W \subset T^* M$ pulls back to a constant rank subbundle $f^* W \subset T^* N$, as we see from those local coordinates, by pulling back sections. Hence $f^* W$ has the form $f^* W=T^{\perp} f^*F$ for a unique smooth foliation $f^*F$ just when $W$ has the form $W=T^{\perp}F$ for a smooth foliation $F$.

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