I am trying to prove the following statement
Given a Noetherian scheme $X$ and a surjective morphism of quasi-coherent sheaves on $X$: $$\mathcal{F}\xrightarrow{f} \mathcal{G} \to 0$$ where $\mathcal{G}$ is coherent, we can find a subsheaf $\mathcal{F'}$ of $\mathcal{F}$, which is coherent, such that the induced morphism $$\mathcal{F'}\rightarrow \mathcal{G} \to 0$$ is also surjective.
My thought:
The algebraic version of this is obvious, that is, given a Noetherian ring $A$ and an epimorphism of $A$-modules $$M\to N\to 0$$ where $N$ is finitely generated, we can find a finitely generated submodule $M'$ such that the induced map $M'\to N$ is surjective.
So it is natural to restrict the morphism of sheaves on an affine open subset (we can find a finite affine open cover $\{U_i\}$ of $X$), $$\mathcal{F}|_{U_i}\cong \widetilde{M_i}, \ \ \ \ \mathcal{G}|_{U_i}\cong \widetilde{N_i}$$ for some $A_i$-modules $M_i$ and $N_i$, where $U_i=\mathrm{Spec}(A_i)$.
Then for the map $M_i\to N_i$, we can find the finitely generated submodule $M_i'$. Thus, the coherent sheaf $\widetilde{M_i'}$ is the desired subsheaf, but unfortunately, only on $U_i$.
I am stuck at the point how to glue these subsheaves on $U_i$ together to get a coherent sheaf on $X$.
Any answer and hints are welcome!
I am not sure if the statement is true, but it is a promising "lemma" to show the derived category of the complexes of coherent sheaves is equivalent to the derived category of the complex of quasi-coherent sheaves with coherent cohomology, that is, $$D^b(\mathrm{Coh}(X))\simeq D^b_{\mathrm{Coh}(X)}(\mathrm{QCoh}(X)).$$
Update:
This is true (see Lemma 3.6 on Huybrechts' Fourier-Mukai Transforms in Algebraic Geometry) and his proof is clear.