Splitting of the tangent bundle of a vector bundle

Question

Let $\pi:E\to M$ be a rank $k$ vector bundle over the (compact) manifold $M$ and let $i:M\hookrightarrow E$ denote the zero section. I'm interested in a splitting of $i^*(TE)$, the restriction of the tangent bundle $TE$ to the zero section.

Intuitively I would guess that one could show the following:

$$i^*(TE)\cong TM\oplus E$$

Is this true? If so, how does the proof work?

Any details and references are appreciated!

Answer 1

The morphism $\pi:E\to M$ (which is a submersion) induces a surjective tangent morphism $T\pi: TE\to \pi^*TM\to 0$ whose kernel is (by definition) the vertical tangent bundle $T_vE$ .
There results the exact sequence of bundles on E $$0\to T_vE\to TE\stackrel {T\pi}{\to} \pi^*TM\to 0$$ Pulling back that exact sequence to $M$ via the embedding $i$ yields the exact sequence of vector bundles on M: $$0\to E\to TE\mid M \to TM\to 0 \quad (\bigstar )$$ The hypothesis that $M$ is compact is irrelevant to what precedes.
However if $M$ is paracompact, the displayed sequence $(\bigstar )$ splits and you may write $$TE\mid M \cong E\oplus TM$$
Since however the splitting of $(\bigstar)$ is not canonical, I do not recommend this transformation of the preferable (because intrinsic) exact sequence$(\bigstar)$.

Edit ( September 22, 2016)
I forgot to mention the interesting fact that $T_vE$ and $\pi^*E$ are canonically isomorphic as vector bundles on $E$, so that we have a canonical exact sequence on $E$: $$0\to \pi^*E\to TE\stackrel {T\pi}{\to} \pi^*TM\to 0 $$The equality $T_vE=\pi^*E$ ultimately rests on the fact that the tangent space to a vector space $V$ at any point $v\in V$ is canonically isomorphic to that vector space : $T_vV=V$.

Answer 2

Just make a comment what have shown above we can use this exact sequence to compute differential of vector bundle morphism :

Let $E\to M$ , $F\to M$ be 2 smooth vector bundles, and $f : E\to F$ be a smooth vector bundle morphism.We know that we have exact sequence

$$0\to E\to TE|_M\to TM\to 0\\0\to F\to TF|_M\to TM\to 0$$ and bundle morphism $f, df|_M, id$ provide a morphism between these two complex, the second square commute since $\pi_E = \pi_F \circ f$, therefore, $d\pi_E = d\pi_F \circ df$, the first square commute as $E\to TE|_M$ simply as the inclusion of zero section and $df$ will maps zero to zero as it's $\Bbb{R}-$linear.

Therefore the map $df|M : TE|_M \to TF|_M$ under the identication above is just the map $$df|_M:TM\oplus E \to TM\oplus F\\(v,e)\mapsto (v,f(e))$$