InsideDarkWeb.com

Are final functors stable under pullback?

Recall the notion of a final functor, which is a sort of colimit-preservation property.
Is such class of functors stable under pullbacks in Cat? Namely, is the pullback of a final functor along any other functor still final? If not, what is a counterexample?

A reference would also be welcome.

Mathematics Asked by geodude on November 12, 2021

1 Answers

One Answer

No, final functors are not stable under pullback in general.

Let $I := {0to1}$ be the walking arrow category, then $1:*to I$ picks out the terminal object and is thus final. The diagram $require{AMScd}$ begin{CD} varnothing @>>> *\ @V{F}VV @VV{1}V\ * @>>{0}> I end{CD} is a pullback square, but the functor $F:varnothingto*$ is not final. Indeed, let $G:*tomathbf{Set}$ pick any nonempty set $X$, then $varinjlim G=X$, but $varinjlim Gcirc F=varnothing$ since the colimit of the empty diagram is just the initial object in $mathbf{Set}$. As the unique map $varnothingto X$ is not bijective, we can conclude that $F$ is not final despite the finality of $1:*to I$.

However, final functors form an orthogonal factorisation system with discrete fibrations, and so in particular are closed under pushouts in $mathbf{Cat}$ (see e.g., here)

Answered by shibai on November 12, 2021

Add your own answers!

Related Questions

Decision tree root

0  Asked on February 17, 2021 by john-l

   

Convert Hexadecimal $8D(16)$ to binary in signed magnitude

1  Asked on February 17, 2021 by zamir-ibrahim

   

Most General Unifier computation

1  Asked on February 16, 2021 by milano

   

Ask a Question

Get help from others!

© 2021 InsideDarkWeb.com. All rights reserved.