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Stream: learning: questions

Topic: local universe model: pullback stability

Nico Beck (Oct 10 2023 at 20:01):

The "local universe" splitting is the left adjoint splitting $L:\operatorname{Fib}_{\mathscr S}\to [\mathscr S^{op},\operatorname{Cat}]$ of a Grothendieck fibration. Does somebody happen to know if $L$ is compatible with base change? When $F:\mathscr S'\to \mathscr S$ is a functor, will $L(F^\ast C)$ be isomorphic to $F^\ast (LC)$?

Mike Shulman (Oct 10 2023 at 21:30):

Nope. (-:

Nico Beck (Oct 10 2023 at 21:43):

Mike Shulman said:

Nope. (-:

They won't be isomorphic?

Nico Beck (Oct 10 2023 at 21:45):

This is very unconvenient :/

Mike Shulman (Oct 10 2023 at 22:19):

Think about it: an object of $L(F^*C)$ in the fiber over $x\in \mathcal{S}'$ is an object $y\in \mathcal{S}'$, a morphism $x\to y$, and an object $a\in C(F(y))$. While an object of $F^*(LC)$ in the fiber over $x$ is an object of $LC$ in the fiber over $F(x)$, which is an object $z\in \mathcal{S}$, a morphism $f(x)\to z$, and an object $b\in C(z)$.

Nico Beck (Oct 10 2023 at 23:17):

Mike Shulman said:

Think about it: an object of $L(F^*C)$ in the fiber over $x\in \mathcal{S}'$ is an object $y\in \mathcal{S}'$, a morphism $x\to y$, and an object $a\in C(F(y))$. While an object of $F^*(LC)$ in the fiber over $x$ is an object of $LC$ in the fiber over $F(x)$, which is an object $z\in \mathcal{S}$, a morphism $f(x)\to z$, and an object $b\in C(z)$.

You are right, the fibers look very different. Is it possible that I have more luck with the right adjoint splitting $\operatorname{Sp}$, at least for reindexings along functors $\mathscr S/\Delta\to \mathscr S/\Gamma$ induced by morphisms $u:\Delta \to \Gamma$ in the base? Because if $p:C\to \mathscr S/\Gamma$ is a fibration and $v:\Theta \to \Delta$ is an object in $\mathscr S/\Delta$, then $\operatorname{Sp}(u^\ast p)(v) = Fib_{\mathscr S/\Delta}(y(v),u^\ast p)$ while $u^\ast (\operatorname{Sp} p)(v) = (\operatorname{Sp} p)(uv) = Fib_{\mathscr S/\Gamma}(y(uv),p)$. If I did no mistake then I should get the picture below
Screenshot-2023-10-11-011414.png
I feel like that I should get $Fib_{\mathscr S/\Delta}(y(v),u^\ast p) \cong Fib_{\mathscr S/\Gamma}(y(uv),p)$ from that, but I am not exactly sure.

Mike Shulman (Oct 11 2023 at 01:30):

If $R$ denotes the right adjoint splitting, then an object of $R(F^*C)$ in the fiber over $x$ consists of, for every $y$ and morphism $y\to x$, an object of in the fiber of $C$ over $F(y)$, plus some isomorphisms. Whereas an object of $F^*(RC)$ in the fiber over $x$ is an object in the fiber of $RC$ over $F(x)$, which consists of, for every $z$ and morphism $z\to F(x)$, an object in the fiber of $C$ over $z$, plus some isomorphisms. They don't look the same to me, and I don't see why $F$ being the induced functor between a pair of slice categories should matter.

Nico Beck (Oct 11 2023 at 06:15):

Mike Shulman said:

If $R$ denotes the right adjoint splitting, then an object of $R(F^*C)$ in the fiber over $x$ consists of, for every $y$ and morphism $y\to x$, an object of in the fiber of $C$ over $F(y)$, plus some isomorphisms. Whereas an object of $F^*(RC)$ in the fiber over $x$ is an object in the fiber of $RC$ over $F(x)$, which consists of, for every $z$ and morphism $z\to F(x)$, an object in the fiber of $C$ over $z$, plus some isomorphisms. They don't look the same to me, and I don't see why $F$ being the induced functor between a pair of slice categories should matter.

It should matter because $(\mathscr S/\Delta)/v=(\mathscr S/\Theta)$ so that the (morphisms with codomain $v$ in $\mathscr S/\Delta$) = (morphisms with codomain $\Theta$ in $\mathscr S$) = (morphisms with codomain $uv$ in $\mathscr S/\Gamma$).
This means that an object in the $(v:\Theta \to \Delta)$ th fiber of $Ru^\ast C$ consists of an object in $C(uv)$ plus for every $w: \Xi \to \Theta$ an object in $C(uvw)$ plus some isomorphisms. Similarly an object in the $(v:\Theta \to \Delta)$ th fiber of $u^\ast RC$ consists of an object in $(RC)(uv)$ which is the same thing as an object in $C(uv)$ plus for each morphism $w:\Xi \to \Theta$ some object in $C(uvw)$ plus some isomorphisms. So the same data on both sides?

Mike Shulman (Oct 11 2023 at 09:38):

Ah, I see what you're saying. That seems possible, though I haven't checked the details.