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## Main Question or Discussion Point

Say one one have a projection map ##\pi : M^5 \to M^4## which in adapted coordinates are of the form

$$ \pi(x^\mu, x^4) = x^\mu$$

where ##\mu = 0,1,2,3##. Now if one ##M^4## introduce an orthonormal frame ##\left\{ e_\mu, e_4\right\}## where ##e_\mu## are tangential to ##M^4## and ##e_4## orthogonal to it. The corresponding dual basis is ##\left\{\omega^\mu, \omega^4\right\}##. If one now considers ##\omega^\nu## to be part of the cotangent space ##T^*_pM^4## and one take the exterior derivative

$$d ^{^{(4)}}\omega^\nu$$

Where ##^{^{(4)}}\omega^\nu## denotes that i consider the covector a part of ##T_p^*M^4##. Now if one considers the same covector as a covector in ##T^*_pM^5## and we take the exterior derivative

$$d ^{^{(5)}}\omega^\nu$$

would it then be correct to say that

$$\pi^* \left( d ^{^{(4)}}\omega^\nu\right) = d \pi^* \left(^{^{(4)}}\omega^\nu\right) = d^{^{(5)}}\omega^\nu$$

due to the commutation of the pullback ##\pi^*## with the exterior derivative?

$$ \pi(x^\mu, x^4) = x^\mu$$

where ##\mu = 0,1,2,3##. Now if one ##M^4## introduce an orthonormal frame ##\left\{ e_\mu, e_4\right\}## where ##e_\mu## are tangential to ##M^4## and ##e_4## orthogonal to it. The corresponding dual basis is ##\left\{\omega^\mu, \omega^4\right\}##. If one now considers ##\omega^\nu## to be part of the cotangent space ##T^*_pM^4## and one take the exterior derivative

$$d ^{^{(4)}}\omega^\nu$$

Where ##^{^{(4)}}\omega^\nu## denotes that i consider the covector a part of ##T_p^*M^4##. Now if one considers the same covector as a covector in ##T^*_pM^5## and we take the exterior derivative

$$d ^{^{(5)}}\omega^\nu$$

would it then be correct to say that

$$\pi^* \left( d ^{^{(4)}}\omega^\nu\right) = d \pi^* \left(^{^{(4)}}\omega^\nu\right) = d^{^{(5)}}\omega^\nu$$

due to the commutation of the pullback ##\pi^*## with the exterior derivative?