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 $1$.



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> **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?
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Does every bicategory have a "delaxing object"?
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 $\mathfrak A \to \mathfrak B$.


