Lemma 95.4.1. A morphism $(f, \varphi ) : (Y, \mathcal{G}) \to (X, \mathcal{F})$ of $\mathcal{QC}\! \mathit{oh}$ is strongly cartesian if and only if the map $\varphi $ induces an isomorphism $f^*\mathcal{F} \to \mathcal{G}$.
Proof. Let $(X, \mathcal{F}) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{QC}\! \mathit{oh})$. Let $f : Y \to X$ be a morphism of $(\mathit{Sch}/S)_{fppf}$. Note that there is a canonical $f$-map $c : \mathcal{F} \to f^*\mathcal{F}$ and hence we get a morphism $(f, c) : (Y, f^*\mathcal{F}) \to (X, \mathcal{F})$. We claim that $(f, c)$ is strongly cartesian. Namely, for any object $(Z, \mathcal{H})$ of $\mathcal{QC}\! \mathit{oh}$ we have
\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{QC}\! \mathit{oh}}((Z, \mathcal{H}), (Y, f^*\mathcal{F})) & = \coprod \nolimits _{g \in \mathop{\mathrm{Mor}}\nolimits _ S(Z, Y)} \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(\mathcal{O}_ Z)}(g^*f^*\mathcal{F}, \mathcal{H}) \\ & = \coprod \nolimits _{g \in \mathop{\mathrm{Mor}}\nolimits _ S(Z, Y)} \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(\mathcal{O}_ Z)}((f \circ g)^*\mathcal{F}, \mathcal{H}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{QC}\! \mathit{oh}}((Z, \mathcal{H}), (X, \mathcal{F})) \times _{\mathop{\mathrm{Mor}}\nolimits _ S(Z, X)} \mathop{\mathrm{Mor}}\nolimits _ S(Z, Y) \end{align*}
where we have used Equation (95.4.0.1) twice. This proves that the condition of Categories, Definition 4.33.1 holds for $(f, c)$, and hence our claim is true. Now by Categories, Lemma 4.33.2 we see that isomorphisms are strongly cartesian and compositions of strongly cartesian morphisms are strongly cartesian which proves the “if” part of the lemma. For the converse, note that given $(X, \mathcal{F})$ and $f : Y \to X$, if there exists a strongly cartesian morphism lifting $f$ with target $(X, \mathcal{F})$ then it has to be isomorphic to $(f, c)$ (see discussion following Categories, Definition 4.33.1). Hence the "only if" part of the lemma holds. $\square$
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