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I've wondered for a while now if for vector bundles $A\to E$, $B\to E$, is $\Gamma(A\oplus B)\cong \Gamma(A)\oplus\Gamma(B)$? I found a related question here, on this question for tensor products, but I cannot find an answer for the direct sum version.

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user2520938
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  • Unlike the tensor product case, the inclusion $A \to A\oplus B$ and projection $A\oplus B \to A$ (and similarly for $B$) induce maps between $\Gamma(A\oplus B)$ and $\Gamma(A)\oplus \Gamma(B)$. (Also, $E$ is usually used for the total space of a bundle, so using it as the base space here is totally fine but slightly odd.) – anomaly Apr 20 '16 at 15:17
  • Yes. More generally, taking global sections preserves all limits, and in particular all products. – Qiaochu Yuan Apr 20 '16 at 15:52
  • @QiaochuYuan Could you point me to some reference where this result is discussed? Since I have no idea how to prove this. – user2520938 Apr 20 '16 at 19:17
  • @anomaly So it seems that in the tensor case we still have maps $A\to A\otimes B$ (same for $B$) that induce $\Gamma(A)\otimes \Gamma(B)\to \Gamma(A\otimes B)$, but we have no natural map $A\otimes B\to A$, and so we dont have the other map, is that correct? – user2520938 Apr 20 '16 at 19:53
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    Right, exactly. – anomaly Apr 20 '16 at 20:29
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    @user2520938: taking global sections means taking $\text{Hom}(1, -)$ where $1$ is the trivial $1$-dimensional bundle. Hom always preserves colimits in the first variable and limits in the second variable. – Qiaochu Yuan Apr 20 '16 at 20:32

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