I am trying to understand the proof of Theorem 4.6.2.1 in the book on algebraic stacks by Laumon and Moret-Bailly. The setting is this: $S$ is a Noetherian scheme, $f\colon X \rightarrow S$ is a projective $S$-scheme such that $f_*\mathscr{O}_X = \mathscr{O}_S$ universally (i.e. also after any base change on $S$), and we are trying to prove that the stack $\mathscr{C}oh_{X/S}$ that parametrizes $U$-flat quasi-coherent locally of finite presentation $\mathscr{O}_{X_{U}}$-modules for variable $S$-schemes $U$ is actually algebraic.

The proof proceeds by arguing that $$ \bigsqcup_{N, n \ge 0} \mathrm{Quot}^{\circ}_{\mathscr{O}_X^N/X/S} \rightarrow \mathscr{C}oh_{X/S},$$ is a smooth atlas, where for a variable $S$-scheme $U$ the $S$-scheme $\mathrm{Quot}^{\circ}_{\mathscr{O}_X^N/X/S}$ parametrizes $S$-flat $\mathscr{O}_{X_{U}}$-module quotients $\mathscr{Q}$: $$(\alpha: \mathscr{O}_{X_U}^N \twoheadrightarrow \mathscr{Q})$$ satisfying

1) $R^p(f_U)_* \mathscr{Q} = 0$ for $p > 0$,

2) $(f_U)_*(\alpha)\colon \mathscr{O}_U^N \rightarrow (f_U)_*(\mathscr{Q})$ is an isomorphism (in particular, $(f_U)_*(\mathscr{Q})$ is locally free of rank $N$).

The atlas map is $\alpha \mapsto \mathscr{Q}(-n)$ (the same $n$ that is part of the index of the disjoint union).

The proof concludes by arguing surjectivity of the atlas map: take any $U \rightarrow \mathscr{C}oh_{X/S}$ with $U$ affine, it corresponds to an $U$-flat $\mathscr{O}_{X_{U}}$-module $\mathscr{M}$. We need to show that etale locally on $U$ the module $\mathscr{M}$ comes from some $\mathscr{Q}$ and $n$. In fact, Zariski locally will suffice: take any large $n$ for which $R^p(f_U)_*(\mathscr{M}(n)) = 0$ for all $p > 0$ and for which $(f_U)^*((f_U)_*(\mathscr{M}(n))) \rightarrow \mathscr{M}(n)$ is surjective (we use projectivity of $f_U$), then subdivide $U$ into $U_{N, n}$ according to the rank $N$ of the locally free $\mathscr{O}_U$-module $(f_U)_*(\mathscr{M}(n))$, and observe that
$$U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S} \rightarrow U$$
factors through $U_{N, n}$ and is in fact the $\mathrm{GL}_{N, U_{N, n}}$-torsor
$$\mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))).$$
I agree that the map factors through $U_{N, n}$ but the rest is unclear to me: by definition the displayed fiber product should be parametrizing isomorphisms over $X_{U_{N, n}}$, so I presume given an "element" $\beta$ in this $\mathscr{I}som$ functor one realizes $\mathscr{M}(n)$ as a $\mathscr{Q}$ as above via the composition
$$(f_{U_{N, n}})^*(\beta)\colon \mathscr{O}_{X_{U_{N, n}}}^N \xrightarrow{\sim} (f_{U_{N, n}})^*((f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))) \twoheadrightarrow \mathscr{M}_{U_{N, n}}(n).$$
But to check that this is really a valid $\mathscr{Q}$ one **needs to check** 2) above, namely, that $(f_{U_{N, n}})_*((f_{U_{N, n}})^*(\beta))$ is an **isomorphism**. My **question** is: why is this an isomorphism? This being an isomorphism is like saying that a set of global sections generating the sheaf $\mathscr{M}_{U_{N, n}}(n)$ already generates $\Gamma(X_{U_{N, n}}, \mathscr{M}_{U_{N, n}}(n))$, which seems doubtful. By cohomology and base change, in checking this one may assume that the base is a field, but even this case is unclear to me.

I apologize that the question seems so long. The length came from trying to state it in a self-contained way, but in the end this is a basic question about a key step in the proof of the algebraicity of $\mathscr{C}oh_{X/S}$. I would be very grateful if someone could explain how to check the missing compatibility mentioned above (alternatively, how to modify the argument in the book) -- this would clear up any doubts about the completeness of the proof.

EDIT. **Here is a short version of the question summarizing the issue**

*By definition*, $U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S}$ parametrizes a $\mathscr{Q}$ (with its $\alpha$) + an isomorphism $x$ *over $X$* between the $\mathscr{Q}$ and the $\mathscr{M}(n)$. On the other hand, $\mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n)))$ parametrizes isomorphisms $u$ *over (an open of)* $U$. We need to identify the two functors. There are $U$-maps
$$U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S} \rightarrow \mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))), \quad x \mapsto f_*(x\circ \alpha)$$
and
$$\mathscr{I}som(\mathscr{O}_{U_{N, n}}^N, (f_{U_{N, n}})_*(\mathscr{M}_{U_{N, n}}(n))) \rightarrow U \times_{\mathscr{C}oh_{X/S}} \mathrm{Quot}^{\circ}_{\mathscr{O}_{X}^N/X/S}, \quad u \mapsto \mathrm{(adjunction\ surjection)} \circ(f^*u),$$
which are candidates for the sought isomorphism. My question is why is the second of these maps well-defined? In other words, why does the surjection
$$\mathrm{(adjunction\ surjection)}\circ (f^*u) \colon \mathscr{O}_{X_{U_{N, n}}}^N \twoheadrightarrow \mathscr{M}(n)$$
satisfy the condition 2) above (i.e. why is its pushforward an isomorphism)?