Splitting the Tangent Bundle of a Vector Bundle along the Zero Section Good evening everyone, I have a small question:
Assume we have a vector bundle $E = \bigcup\limits_{x\in M} E_x$ over a manifold $M$. I now want to show the following well-known equation: $$TE_M \cong TM \oplus E$$
I've seen it done with short exact sequences, but I wonder if it can be done more elementary: Let $s : M \hookrightarrow E$ be the zero section (i.e. $s(x) = 0 \in E_x$), then we get for each $x\in M$ the direct sum $E_x = \{0\} \oplus E_x = s(x) \oplus E_x$, so:
$$E= s(M) \oplus E$$
For the tangent bundle, this means (identifying $x = s(x)$) $$T_{x}E = T_{x}s(M) \oplus T_{x}E \cong T_xM \oplus E_x,$$ since $s(M) \cong M$ and $T_xE = T_x E_x \cong E_x$ as it is a vector space.
So, my question is: Does anyone see a mistake or something essential missing from my "proof"?
 A: Come to think of it, the above way seems more complicated than I initially thought. I came up with a better way to prove $T_xE = T_x M \oplus E_x$ that also makes clear why we have this particular splitting:
If we denote by $\pi : E \rightarrow M$ the bundle map, i.e. $v \in E_x \mapsto x\in M$, then we can use the differential $d\pi : TE \rightarrow TM$:
$$\ker d\pi(x) = \{ v \in T_{x} E \,\vert\, \exists \gamma \in C^1((-\varepsilon,\varepsilon), E) : \gamma(0) = x, \dot{\gamma}(0) = v, \text{ and }\frac{d}{d t} \pi (\gamma(t)) \vert_{t=0} = 0\}$$
This means $\pi(\gamma(t))$ is constant, so $\gamma(t) \in E_x$ for all $t\in(-\varepsilon,\varepsilon)$, which is equivalent to $v = \dot{\gamma}(0) \in T_x E_x$, so we get $\ker d\pi(x) = T_xE_x \cong E_x$.
On the other hand, since $d\pi$ is surjective (b.c. of $M\hookrightarrow E$), and since it is also linear, it induces the bijection
$$T_xE/T_xE_x\cong T_x M$$
and hence we get the desired formula.
EDIT
I forgot to check that $T_xE_x\cap ds(T_xM) = \{0\}$. This follows because $s$ is the zero-section: If $v\in T_x E_x \cap ds(T_xM)$, there exists a curve $\gamma(t) \in M\, \forall t\in (-\varepsilon,\varepsilon)$ with $v = \frac{d}{dt} s(\gamma(t)) \vert_{t=0}$, but $s(\gamma(t)) \in E_x$ is constantly $0_x \in E_x$, so $v = 0$.
Maybe all this has been obvious from the start, but I'm kinda new to differential topology, so please excuse my initial question.
