I am asked to solve the following exercise.
Let $E = E[M; \pi, \mathbb{R}^n]$,$F = F[M; \pi, \mathbb{R}^m]$, $H = H[M; \pi, \mathbb{R}^k]$ be three smooth vector bundles over $M$ of finite rank $n,m,k \ge 1$. Assume there exists the following short exact sequence of vector bundles \begin{equation*} 0 \longrightarrow E \longrightarrow F \longrightarrow H \longrightarrow 0 \end{equation*} where $\alpha: E \rightarrow F$ and $\beta: F \rightarrow H$ are vectors bundle homomorphisms such that $\beta \circ \alpha = 0$, $\alpha$ being injective and $\beta$ surjective. Prove that, if there exists a vector bundle homomorphism $\lambda: H \rightarrow F$ such that $\beta \circ \lambda = id|_H$, then $F \simeq E \oplus F$, where $E \oplus F$ is the Whitney sum of the two vector bundles.
I reasoned as follows.
Fiberwise (namely, for every $p \in M$), it holds true (using the standard form of the Splitting Lemma for vector spaces) that $F|_p \simeq E|_p \oplus H|_p$, where $\oplus$ here denotes the direct sum of the vector spaces $E|_p$ and $H|_p$. My idea was to consider an open cover of $M$, $\{U_{\alpha}\}_{\alpha \in I}$, where $I$ is a countable index set, such that $p \in U$, $F|_{U} \simeq U \times \mathbb{R}^m$ and also $E|_{U} \simeq U \times \mathbb{R}^n$, $H|_{U} \simeq U \times \mathbb{R}^k$(dropping the subscript for notational ease). Then, we can define \begin{equation*} \Phi : E|_{U} \oplus H|_{U} \longrightarrow F|_{U}, (p, (e,h)) \longmapsto (p, \alpha(e) + \lambda(h)) \end{equation*} Evidently, $\Phi|_q$ is linear for every $q \in U$, being a vector space isomorphism. In addition
- $\Phi$ is injective. In fact, $\Phi(p, (e,h))= (p, \alpha(e) + \lambda(h)) = 0$ implies that $\alpha(e) = 0 \Rightarrow e=0$ since $\alpha$ is injective and $\lambda(h) = 0 \Rightarrow \beta \circ \lambda (h) = h = 0$.
- $\Phi$ is surjective. $\forall (p,f) \in F|_{U}$, it suffices to choose $e = 0$ and $h = \beta(f)$ in order to get $\Phi(p,(0, \beta(f))) = (p,0+\lambda \circ \beta f) = (p,f)$.
I am not sure on how to proceed to demonstrate the smoothness of $\Phi$. My first thought was that it could follow from the fact that both $\alpha$ and $\lambda$ are vector bundle homomorphism.
Does this proves the statement or have I left something out? If anyone could give a check on and correct the above reasoning, it would be greatly appreciated.