$T(F' \circ F) = TF' \circ TF$

Question

Let $M, N,P$ be smooth manifolds and $F : M → N, F' : N \to P$ are smooth maps. We know that there are corresponding smooth maps $TF : TM \to TN, TF': TN \to TP$, where $TM$ denotes the tangent bundle.

Now $F' \circ F : M \to P$ is a smooth map.

From here can we say that $T(F' \circ F) = TF' \circ TF$?

Is the above reason correct?

Answer 1

Maybe we can go for the proof following the definition of pushforward as well?

It suffices to show that $\forall\,p\in M$, $\forall\,X_p\in T_pM$, $\forall\,f\in C_{F'\circ F(p)}^{\infty}(P)$, $$ \left\{\left[T(F'\circ F)\right]_p(X_p)\right\}(f)=\left[TF'_{F(p)}\circ TF_p(X_p)\right](f). $$ Provided the smoothness of $F$ and $F'$, we have $f\circ F'\in C_{F(p)}^{\infty}(N)$ and $f\circ F'\circ F\in C_p^{\infty}(M)$. Thus for one thing, $$ \left\{\left[T(F'\circ F)\right]_p(X_p)\right\}(f)=X_p\left[f\circ\left(F'\circ F\right)\right]=X_p(f\circ F'\circ F). $$ For another, \begin{align} \left[TF'_{F(p)}\circ TF_p(X_p)\right](f)&=\left\{TF'_{F(p)}\left[TF_p(X_p)\right]\right\}(f)\\ &=\left[TF_p(X_p)\right](f\circ F')\\ &=X_p\left[(f\circ F')\circ F\right]=X_p(f\circ F'\circ F). \end{align} Thanks to the arbitrariness of $p$, $X_p$ and $f$, the above two results lead to $$ T(F'\circ F)=TF'\circ TF. $$