How exact is the functor "tensoring with a locally free sheaf"

Solution 1:

Just to remove this question from the list of unanswered questions. Let $$0\to \mathcal A_1\to\mathcal A_2\to\mathcal A_3\to 0$$ be an exact sequence of $\mathcal O_X$-modules and let $\mathcal F$ be a locally free sheaf of rank $n$. Then for any $x\in X$ we have $\mathcal F_x\cong\mathcal O_x^{\oplus n}$, therefore $(\mathcal A_i\otimes\mathcal F)_x\cong(\mathcal A_i)_x^{\oplus n}$. The sequence $$0\to (\mathcal A_1)_x^{\oplus n}\to(\mathcal A_2)_x^{\oplus n}\to(\mathcal A_3)_x^{\oplus n}\to 0$$ is exact, i.e. the sequence $$0\to (\mathcal A_1\otimes\mathcal F)_x\to(\mathcal A_2\otimes\mathcal F)_x \to (\mathcal A_3\otimes\mathcal F)_x\to 0$$ is exact. The sequence $$0\to \mathcal A_1\otimes\mathcal F\to \mathcal A_2\otimes\mathcal F \to \mathcal A_3\otimes\mathcal F \to 0$$ is exact on stalks, therefore it is exact.