Let $X$ be an affine scheme. Suppose we have an exact sequence of $\mathcal{O}_{X}$-modules \begin{equation*} 0 \to \mathcal{F}_{1} \xrightarrow{\phi} \mathcal{F}_{2} \xrightarrow{\psi} \mathcal{F}_{3} \to 0. \end{equation*} If $\mathcal{F}_{2}$ is quasi-coherent, does this imply that the sequence of global sections \begin{equation*} 0 \to \mathcal{F}_{1}(X) \xrightarrow{\phi(X)} \mathcal{F}_{2}(X) \xrightarrow{\psi(X)} \mathcal{F}_{3}(X) \to 0 \end{equation*} is also exact?
If we assume that $\mathcal{F}_{1}$ is quasi-coherent, the answer is yes. This is proved in Hartshorne, Proposition II.5.6. In the proof, the quasi-coherence of $\mathcal{F}_{1}$ is used to show that for a distinguished open affine $D_{f}$ of $X$ and a section $s \in \mathcal{F}_{2}(D_{f})$ there is some positive integer $N$ such that $f^{N}s$ is the restriction of a global section of $\mathcal{F}_{2}$. However, the quasi-coherence of $\mathcal{F}_{2}$ should allow us to conclude this lifting property directly. With this change the argument seems to go through without assuming the quasi-coherence of $\mathcal{F}_{1}$.
I can't see any mistakes in the argument. But I also don't have a good reason why the quasi-coherence of $\mathcal{F}_{2}$ should force $H^{1}(X,\mathcal{F}_{1})$ to vanish. Thanks!