Conference topic
This is from discussions at the 2016 West Coast Algebraic Topology Summer School (WCATSS) at The University of Oregon. Thanks to Zijian Yao for explaining the material.
Consider a morphism of schemes $\varphi:S'\to S$ and coherent sheaves $\mathcal F,\mathcal G$ over $S$. Consider also a map of sheaves $f:\mathcal F\to \mathcal G$ and a map $f'$ between the pullbacks of $\mathcal F$ and $\mathcal G$, as described by the diagram below.
There are two natural questions to ask.
Remark: If 1. is true, then $p_1^*(f') = p_2^*(f')$. If the previous statement is an equivalence, then $\varphi$ is a morphism of descent.
Remark: If 2. is true, then there exists $\alpha:p_1^*(\mathcal G') \to p_2^*(\mathcal G')$ such that $\pi_{32}^*(\alpha)\pi_{21}^*(\alpha) = \pi_{31}^*(\alpha)$ and $\pi^*(\Delta) = \alpha$. If the previous statement is an equivalence, then $\varphi$ is effective.
Consider a morphism of schemes $\varphi:S'\to S$ and coherent sheaves $\mathcal F,\mathcal G$ over $S$. Consider also a map of sheaves $f:\mathcal F\to \mathcal G$ and a map $f'$ between the pullbacks of $\mathcal F$ and $\mathcal G$, as described by the diagram below.
There are two natural questions to ask.
- When is $f' = \varphi^*f$?
- If we start with $\mathcal G'$ over $S'$, when is $\mathcal G' = \varphi^*\mathcal G$?
Remark: If 1. is true, then $p_1^*(f') = p_2^*(f')$. If the previous statement is an equivalence, then $\varphi$ is a morphism of descent.
Remark: If 2. is true, then there exists $\alpha:p_1^*(\mathcal G') \to p_2^*(\mathcal G')$ such that $\pi_{32}^*(\alpha)\pi_{21}^*(\alpha) = \pi_{31}^*(\alpha)$ and $\pi^*(\Delta) = \alpha$. If the previous statement is an equivalence, then $\varphi$ is effective.
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