## August 1, 2007

### Solving Transformations for One of the Two n-Functors

#### Posted by Urs Schreiber

Here is a supposedly basic $n$-categorical question which I keep running into, and which doesn’t seem to be addressed in the literature that I am aware of, nor by the people that I talk to:

fix some integer $n$ and some notion of $\left(n+1\right)$-category of $n$-categories. Then

Given two parallel $n$-functors $F$ and $G$, what is the minimum of conditions and extra structure $S$ on a transformation $t:F\to G$ such that the consistency equation for the component map of $t$ may be solved for $F$ in terms of the data given by $G$, $t$ and $S$?

Under which conditions does a morphism of $n$-functors allow to “solve for” one of the two $n$-functors, in terms of the other?

To start with, a sufficient condition is certainly to demand that $t$ is an equivalence, and to let $S=\left\{\overline{t},i,e,\cdots \right\}$ be a specified weak inverse $\overline{t}$ of $t$ with specified (weak) unit $i$ and counit $e$, and, if $n>2$, further structure.

For instance, to give the simplest example, if two 1-functors $F$ and $G$ are related by an invertible natural transformation $t:F\stackrel{\sim }{\to }G$ we can “solve $F$ for $G$ and $t$” in that for any morphisms $a\stackrel{f}{\to }b$ in the domain, we have $F\left(a\right)\stackrel{F\left(f\right)}{\to }F\left(b\right)\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}F\left(a\right)\stackrel{t\left(a\right)}{\to }G\left(a\right)\stackrel{G\left(f\right)}{\to }G\left(b\right)\stackrel{t\left(b{\right)}^{-1}}{\to }F\left(b\right)$ with the right hand side not involving $F$ itself.

But, while demanding $t$ to be an equivalence is sufficient for doing this, it is far from necessary.

For 1-functors, the sufficient and necessary condition is that $t$ has a right inverse.

But already for 2-functors, the situation becomes more interesting. While I am not quite sure about necessity, I think that a sufficient condition for 2-functors which is truly weaker than demanding an equivalence of 2-functors is to demand that $t:F\to G$ fits into a special ambidextrous adjunction: a left and a right adjunction such that the counit of one is the right-inverse of the unit of the other. See definition 3 here.

This allows to “solve $F$ for $G$”, as described on p. 50.

Now, I have come to begin wondering about the analogous question for $n=3$.

More concretely, I am encountering the following issue: I seem to have a morphism of 3-functors, which behaves a lot like one would expect a pseudoadjunction to behave. But there is one crucial difference:

where in the defintion of an adjunction one usually has an identity morphism, I want a “weak” identity morphism: a morphism which is the identity only lax-ly and op-lax-ly, hence one which is a monad and co-monad on its domain.

I went back to Aaron Lauda’s Frobenius algebras and ambidextrous adjunctions, where on pages 16 and following Verity’s notion of pseudoadjunction is recalled, but it seems I am looking for something even weaker than that: the 2-morphisms on p. 16 labeled “$1$” I would like to allow to be just monads and comonads instead of identities.

I’d be grateful for any comments on this.

Posted at August 1, 2007 5:24 PM UTC

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