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Let $p_A:A\longrightarrow M$ and $p_B:B\longrightarrow N$ be vector bundles and $\Phi:A\longrightarrow B$ a vector bundle morphism covering $\phi:M\longrightarrow N$.

For $p\in M$, define $\Phi_p:=\Phi|_{A_p}:A_p\longrightarrow B_{\phi(p)}$ where $A_p:=p_A^{-1}(p)$ and $B_{\phi(p)}:=p_B^{-1}(\phi(p))$.

Is it true that:

$(i)$ $\Phi$ is surjective $\Leftrightarrow$ $\Phi_p$ is surjective for all $p\in M$;

$(ii)$ $\Phi$ is injective $\Leftrightarrow$ $\Phi_p$ is injective for all $p\in M$;

$(iii)$ $\Phi$ is bijective $\Leftrightarrow$ $\Phi_p$ is bijective for all $p\in M$?

If that is not true then:

$(a)$ which implications do hold?

$(b)$ what would be hypothesis on $\Phi$ or $\phi$ to make it true?

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(i) and (iii) are not true since $\phi$ can be not surjective. Example. Take the trivials bundle $R\times M\rightarrow M$ and $R\times (M+M)\rightarrow M+M$. Where $M+M$ is the disjoint copy of two examples of $M$.

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  • $\begingroup$ Ok, thanks. I guess if $\Phi$ is surjective and $\phi$ is injective then $\Phi_p$ is surjective for every $p\in M$, is it really true? $\endgroup$ – PtF Apr 4 '16 at 11:42
  • $\begingroup$ Yes if $\Phi_p$ is surjective for every $p$ and $\phi$ surjective then $\Phi$ is surjective. $\endgroup$ – Tsemo Aristide Apr 4 '16 at 11:44
  • $\begingroup$ Agreed, but that was not what I meant in my previous comment. $\endgroup$ – PtF Apr 4 '16 at 11:46
  • $\begingroup$ try to construct $\phi$ and $\Phi$ with the example in my answer, they will answer your comment. $\endgroup$ – Tsemo Aristide Apr 4 '16 at 11:48
  • $\begingroup$ Suppose $\Phi$ surjective and $\phi$ is injective: $y\in B_{\phi(p)}\Rightarrow \exists x\in A; y=\Phi(x)$. But $\exists p^\prime\in M; x\in A_{p^\prime}\Rightarrow y\in B_{\phi(p)}\cap B_{\phi(p^\prime)}\Rightarrow \phi(p)=\phi(p^\prime)\Rightarrow p=p^\prime\Rightarrow x\in A_{p}$ and therefore $\Phi_p$ is surjective. It seems right to me, what do you think? $\endgroup$ – PtF Apr 4 '16 at 11:57

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