2
$\begingroup$

It pertains to the classical literature on the subject (see e.g. here) the fact that an accessible 2-monad $T$ on a 2-category $\cal K$ induces an adjunction $$ \mathbf{Strict}\text{-}T\text{-}\textbf{Alg}\leftrightarrows \mathbf{Pseudo}\text{-}T\text{-}\textbf{Alg} $$ between strict $T$-algebras and pseudo $T$-algebras. This means that for every morphism $(A,a)\to (B,b)$ of $T$-algebras where the relevant diagrams commute up to invertible 2-cells, there is a new $T$-algebra $(A^\text{s}, a^\text{s})$ with a strict morphism $(A^\text{s}, a^\text{s}) \to (B,b)$.

What does this construction become when $T : {\bf Cat}\to {\bf Cat}$ is the 2-monad that sends a small category $A$ into the category $A^\textbf{2}$ of its arrows (for sure a finitary monad on a locally finitely presentable category)?

$\endgroup$
10
  • $\begingroup$ Three irresponsible blind guesses: either$${\bf Cat}^\textbf{2}\leftrightarrows{\bf Cat}/\textbf{2}$$or$${\bf Cat}^\textbf{2}\leftrightarrows{\bf Cat}//\textbf{2}$$or$${\bf Cat}/\textbf{2}\leftrightarrows{\bf Cat}//\textbf{2}$$(here "//" stands for hypercomma, a. k. a. lax slice) $\endgroup$ May 8, 2017 at 8:46
  • $\begingroup$ A single naive question: are you saying that ${\bf Cat}^2\cong {\bf Cat}/2$? I've never noticed this. $\endgroup$
    – fosco
    May 8, 2017 at 8:58
  • $\begingroup$ Wait, this is not the only problem. What do these three adjunctions denote? No one of the categories in your guess is the category of algebras for the squaring monad :-) $\endgroup$
    – fosco
    May 8, 2017 at 9:02
  • $\begingroup$ Re first - lhs embeds into rhs as fibrations over $\bf2$. Re second - I believe all three admit adjunctions to $\bf Cat$ inducing the $(\_)^{\bf2}$ monad. $\endgroup$ May 8, 2017 at 9:21
  • 1
    $\begingroup$ Just an à propos: obviously "codescent" shall be banned, and replaced by "ascent" :D $\endgroup$ May 10, 2017 at 7:41

1 Answer 1

2
$\begingroup$

A strict algebra of the squaring monad is precisely a strict factorization system. A normal pseudo-algebra (unit holds strictly) is precisely an orthogonal factorization system. See here. The pseudo-algebras will differ by first replacing a morphism by an isomorphic morphism before factoring it.

So that's what the pseudo-algebras look like. You can read off from this a description of the strictified monad.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.