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Theorem 8.8 #92
Theorem 8.8 #92
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For a possible proof of 2.) that doesn't assume the base to be Segal see [BW23, Proposition 6.1.1. Edit: More precisely, the statement referred to is that any covariant family is inner. |
The proof I implemented also does not use that the base is Segal. |
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That looks great!
First, currently you get the following error:
This is because this definition does not actually rely on |
Thanks! I must have gotten confused and removed the wrong instance of |
The merge-base changed after approval.
Introduce the notions of left fibration and inner fibration as right orthogonal maps wrt
{0} ⊂ Δ¹
andΛ ⊂ Δ²
, respectively.Show that the notions of naive left fibration and left fibration agree. In particular, every projection
∑ A, C -> A
of a covariant familyC : A -> U
is a left fibration (regardless of whetherA
is Segal).Use the calculus of left orthogonal shapes to give a slick proof that every left fibration is an inner fibration.
Deduce that for every covariant family
C : A -> U
over a Segal typeA
, the total type∑ C
is also Segal. This addresses Formalise Theorem 8.8 of RS17 paper #12 (dare I say closes?) unless @cesarbm03 (or anyone else) still wants to implement an alternative proof.Notes:
naiveextext
, since it relies onis-right-orthogonal-to-shape-×
, which does; see Refactor is-right-orthogonal-to-shape-× #91.is-naive-left-fibration-iff-is-covariant
could probably be refactored and shortened now, since the only hard part thereof should be a consequence ofextension-weakening
.