Let $A,B,C$ and $T$ be four holomorphic vector bundles on the complex manifold $X$. If I have a short exact sequence $$0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0$$ is it true that the sequence $$0\longrightarrow A\otimes T\longrightarrow B\otimes T\longrightarrow C\otimes T\longrightarrow 0$$ is also exact? I think that the answer is positive because by the shrt exact sequence on the fibers $$0\longrightarrow A_x\longrightarrow B_x\longrightarrow C_x\longrightarrow 0$$ we obtain a short exact sequence $$0\longrightarrow A_x\otimes T_x\longrightarrow B_x\otimes T_x\longrightarrow C_x\otimes T_x\longrightarrow 0.$$ I have only to prove that the the maps $ A\otimes T\longrightarrow B\otimes T, \: B\otimes T\longrightarrow C\otimes T$ are holomorphic. Any helps?
Exact sequence of holomorphic vector bundles
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vector-bundles
exact-sequence
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0Hint: You may as well work in an open set that simultaneously trivializes $A$, $B$, and $T$; the map $A \to B$ is represented by a matrix of (local) holomorphic functions, and the induced map $A \otimes T \to B \otimes T$ is represented by the Kronecker product with an identity matrix. (Similarly for $B \to C$.) – 2017-01-06
1 Answers
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Indeed I know nothing about vector bundles,but that may help.
$0\rightarrow Z/2Z \rightarrow Q$ as $Z$ module is exact,$Z$ is integer ring,Q means rational number field.
However,$0 \rightarrow Z/2Z\otimes Z \rightarrow Z/2Z\otimes Q$as $Z$ module is not exact.
I think $T$ should be flat.In your case is $T$ flat?
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1Your answer is irrelevant since the OP is talking about vector bundle. The fiber is a vector space, so tensoring is always exact since everything is finite-dimensional. – 2017-01-16
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1Well,I know nothing about it.I really approciate your advise. – 2017-01-18
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0You can google "vector bundles" it if you don't know it. – 2017-01-18