Hey, so there's a fact in Scholze's condensed notes that the derived solidification of the reals R is 0. I'm wondering: Is this true for R^discrete as well?
I want to guess that it's not true, but I'm not sure.
@HarryGindi As a discrete abelian group R is just some huge Q-vector space. So the first question is probably understanding what the solidification of Q is
@EdoardoLanari The definition is easy: if $p:C\to D$ is a functor, an arrow $f:x\to y$ in $C$ is cocartesian iff the square $[\operatorname{Map}_C(y,z)\to \operatorname{Map}_C(x,z)]\to [\operatorname{Map}_D(py,pz)\to \operatorname{Map}_D(px,pz)]$ is cartesian. Then $p$ is cocartesian if for every edge $px\to r$ there is a cocartesian lift $x\to \tilde r$ in $C$
(there are a couple of ways of rephrasing the definition, but I'll leave them as an exercise). No idea about a reference though
@DenisNardin Thanks, I am well aware of this formulation. I was looking for the one that is expressed in terms of the existence of a square with invertible horizontal maps towards a "rigid" fibration. I think I saw it somewhere but couldn't find it.
