# How to prove that the tangent map $T\phi$ into the pullback bundle is smooth?

Assume $\phi: M\rightarrow N$ is smooth. Let $\phi^*(TN)$ be the pullback bundle of $TN$ by $\phi$. Define $T\phi:TM\rightarrow \phi^*(TN)$ as follows: $T\phi(m,v)=(m,d\phi_m(v))$.

We also have the corresponding bundle morphism (covering $\phi$), $\bar \phi : \phi^*(TN)\rightarrow TN$, which I think is smooth by the construction of the smooth structure on the pullbak bundle $\phi^*(TN)$.

So, we have a composition of maps which satisfies: $\bar \phi (T\phi)=d\phi$ where $d\phi:TM\rightarrow TN$ is the usual differential map between the corresponding tangent bundles induced by $\phi$.

I know that $d\phi,\bar \phi$ are smooth. How can I show $T\phi$ is smooth?

I would also be happy to find some reference on this subject (I am reading Lee's Intro' to smooth manifolds, and he doesn't treat pullbacks of vector bundles).

$\require{AMScd}$ \begin{CD} TM @>T\phi>> \phi^*(TN)\ @> \bar \phi>> TN \end{CD}

## 1 Answer

For each smooth vector bundle $$E\to N$$ $$\phi^*(E)$$ is an embedded submanifold of $$M\times E$$. Hence as

$$\require{AMScd}$$ $$\begin{CD} TM @>T\phi>> \phi^*(TN)\ @> inc>> M\times TN \end{CD}$$

is smooth so is $$T\phi$$.