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, MartΓ­n HΓΆtzel EscardΓ³ wrote: > > Let P be a subsingleton and 𝓀 be a universe, and consider the > product map > > Ξ  : (P β†’ 𝓀) β†’ 𝓀 > A ↦ Ξ  (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=πŸ™ (empty or > singleton type). > > But the reasons are fundamentally different: > > (0) If P=𝟘, the domain of Ξ  is equivalent to πŸ™, and Ξ  amounts to > the map πŸ™ β†’ 𝓀 with constant value πŸ™. > > In general, a function πŸ™ β†’ X into a type X is *not* an > embedding. Such a function is an embedding iff it maps the > point of πŸ™ to a point x:X such that the type x=x is a > singleton. > > And indeed for X:=𝓀 we have that the type πŸ™=πŸ™ is a singleton. > > (1) If P=πŸ™, the domain of Ξ  is equivalent to 𝓀, and Ξ  amounts to > the identity map 𝓀 β†’ 𝓀, which, being an equivalence, is an > embedding. > > Question. Is there a uniform proof that Ξ  as above for P a > subsingleton is an embedding, without considering the case > distinction (P=𝟘)+(P=πŸ™)? > > Martin > > -- 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.