Re: [HoTT] Re: Proof that something is an embedding without assuming excluded middle?

[Michael Shulman <shulman@sandiego.edu>]

Here's a sketch of a more conceptual argument.  Both 𝓤 and P → 𝓤 are
the object-types of (oo-)categories, and Π is the object-map of a
right adjoint functor whose counit is an equivalence.  Thus, by a
standard argument, it is fully faithful, and hence also fully faithful
on equivalences (which, by univalence, are the equalities).  Of course
we can't define the whole oo-categories in Book HoTT, but I think this
is one of those arguments that only needs the 1- or 2-dimensional
structure.
On Wed, Nov 14, 2018 at 3:07 AM Paolo Capriotti <p.capriotti@gmail.com> wrote:
>
> On Tue, 13 Nov 2018, at 8:32 PM, Martín Hötzel Escardó wrote:
> > Let P be a subsingleton and 𝓤 be a universe, and consider the
> > product map
> >
> >   Π : (P → 𝓤) → 𝓤
> >          A     ↦ Π (p:P), A(p).
> >
> > Is this an embedding? (In the sense of having subsingleton
> > fibers.)
> >
> > It is easy to see that this is the case if P=𝟘 or P=𝟙 (empty or
> > singleton type).
> >
> > But the reasons are fundamentally different:
> >
> > (0) If P=𝟘, the domain of Π is equivalent to 𝟙, and Π amounts to
> >     the map 𝟙 → 𝓤 with constant value 𝟙.
> >
> >     In general, a function 𝟙 → X into a type X is *not* an
> >     embedding. Such a function is an embedding iff it maps the
> >     point of 𝟙 to a point x:X such that the type x=x is a
> >     singleton.
> >
> >     And indeed for X:=𝓤 we have that the type 𝟙=𝟙 is a singleton.
> >
> > (1) If P=𝟙, the domain of Π is equivalent to 𝓤, and Π amounts to
> >     the identity map 𝓤 → 𝓤, which, being an equivalence, is an
> >     embedding.
> >
> > Question. Is there a uniform proof that Π as above for P a
> > subsingleton is an embedding, without considering the case
> > distinction (P=𝟘)+(P=𝟙)?
>
> I think one can show that ap Π is an equivalence by giving an inverse. Given X, Y : P → 𝓤, first note that the equality type X = Y is equivalent by function extensionality to (u : P) → X u = Y u. Since one can prove easily that (u : P) → X u = Π X, it follows that X = Y is equivalent to P → Π X = Π Y. Hence we get a map ω : Π X = Π Y → X = Y. To show that it is indeed an inverse of ap Π, start with some α : Π X = Π Y. By following the construction above, we get that ap Π (ω α) maps a function h : Π X to λ u . α (λ v . tr^X [u,v] (h u)) u, where [u,v] : u = v is given by the fact that P is a proposition. Now, within the assumption u : P, the λ expression in brackets is equal to h itself, hence ap Π (ω α) = α. The other composition is easier, since it can just be checked on reflexivity.
>
> Best,
> Paolo
>
> --
> You received this message because you are subscribed to the Google Groups "Homotopy Type Theory" group.
> To unsubscribe from this group and stop receiving emails from it, send an email to HomotopyTypeTheory+unsubscribe@googlegroups.com.
> For more options, visit https://groups.google.com/d/optout.

-- 
You received this message because you are subscribed to the Google Groups "Homotopy Type Theory" group.
To unsubscribe from this group and stop receiving emails from it, send an email to HomotopyTypeTheory+unsubscribe@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.