The Stacks project

Lemma 105.16.3. Notation as in Lemma 105.16.2. Assume $\mathcal{X}$ is Noetherian and $g$ is surjective and flat. Then $B \to C$ is universally injective.

Proof. Consider the natural map $1 : \mathcal{O}_\mathcal {X} \to g_{\mathit{QCoh}, *}\mathcal{O}_ U$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. Pulling back to $U$ and using adjunction we find that the composition

\[ \mathcal{O}_ U = g^*\mathcal{O}_\mathcal {X} \xrightarrow {g^*1} g^*g_{\mathit{QCoh}, *}\mathcal{O}_ U \to \mathcal{O}_ U \]

is the identity in $\mathit{QCoh}(\mathcal{O}_ U)$. Write $g_{\mathit{QCoh}, *}\mathcal{O}_ U = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as a filtered colimit of coherent $\mathcal{O}_\mathcal {X}$-modules, see Cohomology of Stacks, Lemma 102.18.1. For $i$ large enough the map $1 : \mathcal{O}_\mathcal {X} \to g_{\mathit{QCoh}, *}\mathcal{O}_ U$ factors through $\mathcal{F}_ i$, see Cohomology of Stacks, Lemma 102.13.5. Say $s : \mathcal{O}_\mathcal {X} \to \mathcal{F}_ i$ is the factorization. Then

\[ \mathcal{O}_ U \xrightarrow {g^*s} g^*\mathcal{F}_ i \to g^*g_{\mathit{QCoh}, *}\mathcal{O}_ U \to \mathcal{O}_ U \]

is the identity. In other words, we see that $s$ becomes the inclusion of a direct summand upon pullback to $U$. Set $\mathcal{F}_ i^\vee = hom(\mathcal{F}_ i, \mathcal{O}_\mathcal {X})$ with notation as in Cohomology of Stacks, Lemma 102.10.8. In particular there is an evaluation map $ev : \mathcal{F}_ i \otimes _{\mathcal{O}_\mathcal {X}} \mathcal{F}_ i^\vee \to \mathcal{O}_\mathcal {X}$. Evaluation at $s$ defines a map $s^\vee : \mathcal{F}_ i^\vee \to \mathcal{O}_\mathcal {X}$. Dual to the statement about $s$ we see that $g^*(s^\vee )$ is surjective, see see Cohomology of Stacks, Section 102.12 for compatibility of $hom$ and $\otimes $ with restriction to $U$. Since $g$ is surjective and flat, we conclude that $s^\vee $ is surjective (see locus citatus). Since $F$ is right exact, we conclude that $F(\mathcal{F}_ i^\vee ) \to F(\mathcal{O}_\mathcal {X}) = B$ is surjective. Choose $\lambda \in F(\mathcal{F}_ i^\vee )$ mapping to $1 \in B$. Denote $e = F(s)(1) \in F(\mathcal{F}_ i)$ the image of $1$ by the map $F(s) : B = F(\mathcal{O}_\mathcal {X}) \to F(\mathcal{F}_ i)$. Then the map

\[ F(ev) : F(\mathcal{F}_ i) \otimes _ B F(\mathcal{F}_ i^\vee ) = F(\mathcal{F}_ i \otimes _{\mathcal{O}_\mathcal {X}} \mathcal{F}_ i^\vee ) \longrightarrow F(\mathcal{O}_\mathcal {X}) = B \]

sends $e \otimes \lambda $ to $1$ by construction. Hence the map $B \to F(\mathcal{F}_ i)$, $b \mapsto be$ is universally injective because we have the one-sided inverse $F(\mathcal{F}_ i) \to B$, $\xi \mapsto F(ev)(\xi \otimes \lambda )$. Since this is true for all $i$ large enough we conclude. $\square$


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