If I'm not mistaken, there is a bicategory $\mathsf{Monad}$ given as follows:

  1. Start with the associative operad.
  2. Deloop it to obtain a multicategory.
  3. Adjoin objects and morphisms as necessary to obtain a monoidal category.
  4. Deloop that to obtain a bicategory.

So basically, $\mathsf{Monad}$ is the "double-delooping" of the associative operad.

I think that $\mathsf{Monad}$ has the following special property: given a bicategory $\mathfrak{B}$, a monad in $\mathfrak{B}$ is the same thing as a functor $\mathsf{Monad} \rightarrow \mathfrak{B}$. Now recall that lax functors $1 \rightarrow \mathfrak{B}$ are the "same as" monads in $\mathfrak{B}$. If so, then it would seem that lax functors out of $1$ are the same as ordinary functors out of $\mathsf{Monad}$. We might call $\mathsf{Monad}$ the "delaxing object" of $1$.

Question. Have I made any mistakes above? If not, is it true that every bicategory $\mathfrak{A}$ has a "delaxing object" $\mathfrak{A}'$ such that lax functors out of $\mathfrak{A}$ are the "same as" ordinary functors out of $\mathfrak{A}'$? If not, is there a way to repair this problem, say by working instead with $(\infty,\infty)$-categories instead of bicategories, or something like that?


2 Answers 2


I think that the theory of lax functors of $(\infty,\infty)$-categories is not sufficiently developed to answer that version of your question. But for your main question, yes.

Given a bicategory $\mathfrak A$, you certainly have a set of objects and for each pair of objects $a,b\in\mathfrak A$ a category $\mathfrak A(a,b)$. Define a new category $\mathfrak A''$ as the free category with the same objects as in $\mathfrak A$ and generated by the categories $\mathfrak A(a,b)$. So for instance a 1-morphism in $\mathfrak A''$ is a composable sequence of $1$-morphisms in $\mathfrak A$. Now given a 1-morphism in $\mathfrak A$, there's a corresponding "generating" 1-morphism in $\mathfrak A''$, and on the other hand given a 1-morphism in $\mathfrak A''$, think of it as a sequence in $\mathfrak A$, and compose, and get a 1-morphism in $\mathfrak A$. Combining these, you find that for every 1-morphism $(f_1,f_2,\dots,f_n)$ in $\mathfrak A''$ there's a corresponding generating 1-morphism $f_1f_2\dots f_n$ in $\mathfrak A''$. (Choose some way of parenthesizing the composition. The choice won't matter because of later imposition of naturality conditions.) This assignment is functorial for the 2-morphisms. Let's now add to $\mathfrak A''$ some new 2-morphisms $(f_1,f_2,\dots,f_n) \Rightarrow f_1f_2\cdots f_n$. Now I want to impose some relations on these new 2-morphisms (and the old ones). First, I'd like them to be "natural" in a sense that I won't unpack, but the idea is that the 2-morphism $(f_1,f_2,\dots,f_n) \Rightarrow f_1f_2\cdots f_n$ depends naturally on the choice of $f_1,\dots,f_n$.

Second, and more importantly, I want that compositions of these new morphisms should agree. Really what I mean is the following. Suppose that you have some monotonic function $\phi:\{1,\dots,n\} \to \{1,\dots,m\}$. Then for each $i\in \{1,\dots,m\}$, $\phi^{-1}(i)$ is an interval $\{j,j+1,\dots,j+k\}$. So given a length-$n$ 1-morphism $(f_1,\dots,f_n)$ in $A''$, it makes sense to compose $f_j\dots f_{j+k}$; as $i$ varies, you get a length-$m$ 1-morphism $(f_1f_2\cdots f_{k_1},f_{k_1+1}f_{k_1+2}\cdots f_{k_2},\dots,f_{k_m}\cdots f_n)$ in $A''$. (What I described earlier is the $m=1$ case. If $\phi^{-1}(i) = \emptyset$, the composition is the appropriate identity map.) What I really want to do is to adjoint to $\mathfrak A''$ a 2-morphism $(f_1,\dots,f_n) \Rightarrow (f_1f_2\cdots f_{k_1},f_{k_1+1}f_{k_1+2}\cdots f_{k_2},\dots,f_{k_m}\cdots f_n)$ for each 1-morphism $(f_1,\dots,f_n)$ in $\mathfrak A''$ and each monotonic function $\phi : \{1,\dots,n\} \to \{1,\dots,m\}$. I demand that these new 2-morphisms be natural for the 2-morphisms I already had, and that moreover the 2-morphism be functorial in $\phi$ in the sense that compositions of monotonic functions correspond to compositions of 2-morphisms (and identity monotonic functions correspond to identity 2-morphisms).

Let me call the resulting bicategory, generated by the 1- and 2-morphisms as above and with the given relations between 2-morphisms, $\mathfrak A'$. If I am not mistaken, ("pseudo") functors $\mathfrak A' \to \mathfrak B$ are the same as lax functors $\mathfrak A \to \mathfrak B$.


Theo's answer gives a very explicit construction; I just wanted to mention that there is an abstract framework for it as well. For any 2-monad $T$, there are notions of strict, pseudo, and lax (and colax) morphisms between $T$-algebras, and if $T$ is well-behaved then the inclusions of the 2-category of strict maps into the other three 2-categories have left adjoints. Bicategories are in fact the algebras of a 2-monad $T$ on the 2-category of "Cat-enriched graphs", and the four kinds of $T$-morphism are the four kinds of functor between bicategories. (The $T$-2-cells are not the usual sort of pseudonatural transformations, though; they are called icons.) Thus, the left adjoint in the lax case (sometimes called the "lax morphism classifier") is what you are looking for. (By definition it gives you that lax morphisms out of $A$ are the same as strict morphisms out of $A'$, but it so happens that $A'$ is always what's called "flexible", so that every pseudo morphism out of it is equivalent to a strict one.) If you unravel the abstract construction of $A'$ as a "codescent object", you'll get Theo's version.


  • Blackwell, Kelly, and Power: Two-dimensional monad theory
  • Lack, Codescent objects and coherence and A 2-categories companion

(By the way, the monoidal category that you delooped to get Monad is the augmented simplex category.)


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