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If I have a short exact sequence of vector bundles \begin{equation} 0\rightarrow E' \rightarrow E \rightarrow E'' \rightarrow 0 \end{equation} then it splits, which mean that $E = E'\oplus E''$? So if I have a short exact sequence of locally free sheaves \begin{equation} 0\rightarrow \mathcal{F}' \rightarrow \mathcal{F}\rightarrow \mathcal{F}'' \rightarrow 0 \end{equation} then is it also true that the sequence splits and I have $\mathcal{F} = \mathcal{F}'\oplus\mathcal{F}''$?

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It depends whether you think of the vector bundles as smooth vector bundles, or as holomorphic vector bundles. Short exact sequences of smooth vector bundles always split (see here). But if our vector bundles are holomorphic vector bundles on complex manifolds and if the morphisms in the SES respect the holomorphic structure of the bundles, then we might want to know whether it's possible to find a splitting where the inclusion/projection morphisms that define the splitting also respect the holomorphic structures.

In general, the answer is no. For example, the Euler sequence on $\mathbb{CP}^n$, $$ 0 \to \mathcal O_{\mathbb P^n} \to \mathcal O_{\mathbb P^n}(1)^{\oplus (n + 1)} \to \mathcal T_{\mathbb P^n} \to 0,$$ splits when viewed as a SES of smooth vector bundles, but it is not possible to find a splitting that respect the holomorphic structure. (See here - I particularly like Ben's answer.)

The second part of your question is about locally free sheaves. On a smooth complex projective variety, locally free sheaves and their morphisms are totally analogous to holomorphic vector bundles and their holomorphic-structure-preserving morphisms on the corresponding complex manifold. So the above Euler sequence serves as a counter-example to your second claim.

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