Let $F:M\rightarrow N$ and $G:N\rightarrow P$ be surface transformation. Show that $(G\circ F)^{*}=F^{*}\circ G^{*}$

Question

$Let F:M\rightarrow N$ and $G:N\rightarrow P$ be surface transformation. Show that $$(G\circ F)^{*}=F^{*}\circ G^{*}$$

Here is definition : Let $F:M\rightarrow N$ transformation of surfaces.

$i$) If $\phi$ is $0$-form then $F^{*}\phi =\phi\circ F$ and it's also $0$-form.

$ii$)If $\phi$ is $1$-form then for every $v$ ($v$ is tangent vector) $F^{*}\phi(v) =\phi(F_{*}(v))$

$iii$) If $\phi$ is $2$-form then for every $v,w$ (both tangent vector) $F^{*}\phi(v,w) =\phi(F_{*}(v),F_{*}(w))$

We should show that $(G\circ F)^{*}=F^{*}\circ G^{*}$ for all cases. Let $\phi$ is $0$-form then $$(G\circ F)^{*}(\phi )=\phi\circ (G\circ F)=(\phi \circ G)\circ F=F^{*}(\phi\circ G)=F^{*}(G^{*}(\phi ))=(F^{*}\circ G^{*})(\phi )$$

But I couldn't show for $1$-form and $2$-form. Any idea will be appreciated.

Answer 1

Here is the proof for $1$-forms, the other cases being similar.

Let $v$ be a vector field on $M$ and $\phi$ be a $1$-form on $P$. Then: \begin{align} \left((G\circ F)^*\phi\right)(v) &= \phi\left((G\circ F)_* v \right) \\ &= \phi\left( \left(G_* \circ F_*\right) v\right) & \text{(chain rule)}\\ &= \phi\left(G_*\left(F_* v\right)\right) \\ &= G^* \left(\phi \right)(F_*v)\\ &= F^* \left(G^*\phi\right)(v)\\ &= \left(F^* \circ G^*\right)(\phi)(v) \end{align}