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Feb 3 at 17:52 comment added Tim Campion Coming back to this, it took me awhile to recall that $B \times_{D'} C \to B' \times_{D'} C \to B' \times_{D'} C' = A'$ is the composite of a base-change of $B \overset \sim \to B'$ and a base-change of $C \overset \sim \to C'$, which is why it's in $\mathcal L$ (as claimed in the first line after the first display).
Apr 12, 2020 at 17:07 vote accept arsmath
Jan 8, 2020 at 21:29 history edited Tim Campion CC BY-SA 4.0
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Jan 7, 2020 at 23:22 history edited Tim Campion CC BY-SA 4.0
added 15 characters in body
Jan 7, 2020 at 23:06 history answered Tim Campion CC BY-SA 4.0