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

["Martín Hötzel Escardó" <escardo.martin@gmail.com>]

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Thanks, Paolo. 

Independently, Mike Shulman had sent me another solution based on modal 
operators off-list. But these two solutions are essentially the same. 

I am glad the answer to the question is positive. 

Maybe I'll explain later why I am interested in this question. 

Best, 
Martin 

On Wednesday, 14 November 2018 11:07:50 UTC, Paolo Capriotti 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
>
>

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