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

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)