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

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

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NB. Of course this map is left-cancellable. But this is weaker than being 
an embedding. M.

On Tuesday, 13 November 2018 20:32:22 UTC, 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)?
>
> Martin
>
>

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