Let $f: X \longrightarrow Y$ be a morphism of smooth projective schemes.
Consider
$$0 \longrightarrow \mathcal{F}_{n} \longrightarrow \cdots \longrightarrow \mathcal{F}_{0} \longrightarrow \mathcal{F} \longrightarrow0$$ the exact sequence of locally free sheaves in $Y$.
It's true that the sequence below is exact?
$$0 \longrightarrow f^{*}\mathcal{F}_{n} \longrightarrow \cdots \longrightarrow f^{*}\mathcal{F}_{0} \longrightarrow f^{*}\mathcal{F} \longrightarrow0$$
Thank you very much.
Best Answer
Yes, even with no assumptions on $f$. Suppose we have an exact sequence $$0\to M_n\to\cdots\to M_1\to M_0\to0$$ of flat $A$-modules. If $N$ is any $A$-module, then the sequence $$0\to M_n\otimes_AN\to\cdots\to M_1\otimes_AN\to M_0\otimes_AN\to0$$ is exact. In the case of $f:\mathop{\mathrm{Spec}}B\to\mathop{\mathrm{Spec}}A$, pullback is given by $-\otimes_AB$, so this applies, and the general case reduces to the affine case.
To prove the claim, break the long exact sequence up into short exact sequences. Starting on the right we have $$0\to K_1\to M_1\to M_0\to0$$ where $K_1=\ker(M_1\to M_0)=\mathrm{coker}(M_3\to M_2)$. Then we get a long exact sequence $$\mathrm{Tor}_1^A(M_0,N)\to K_1\otimes_AN\to M_1\otimes_AN\to M_0\otimes_AN\to0.$$ But $M_0$ is flat, so the Tor vanishes, giving $$0\to K_1\otimes_AN\to M_1\otimes_AN\to M_0\otimes_AN\to0.$$ Because $M_1$ is flat as well, the long exact sequence also shows that $\mathrm{Tor}_1^A(K_1,N)=0$, meaning $K_1$ is also flat. Now look at the next short exact sequence $$0\to K_2\to M_2\to K_1\to0$$ where $K_2=\ker(M_2\to M_1)=\mathrm{coker}(M_4\to M_3)$. We have seen that $K_1$ is flat, so the same argument as above shows that $$0\to K_2\otimes_AN\to M_2\otimes_AN\to K_1\otimes_AN\to0$$ is exact. And so on.