Let $\mathsf{MonCat}_{\mathrm{lax}}$ (resp. $\mathsf{MonCat}_{\mathrm{oplax}}$) be the categories of (say small) not necessarily strict monoidal categories and lax (resp. oplax) monoidal functors. Let $\mathsf{Cat}$ be the category of all small categories. There is an evident forgetful functor $\mathsf{MonCat}_{\mathrm{lax}} \to \mathsf{Cat}$ (resp. $\mathsf{MonCat}_{\mathrm{oplax}} \to \mathsf{Cat}$).

My question: does this functor have a left adjoint?

(I know from Ross-Street's Braided tensor categories that the corresponding forgetful functor from the category of not necessarily strict monoidal categories but with strict monoidal functors to $\mathsf{Cat}$ has a left adjoint. And I wonder if this is still true with lax (or oplax) monoidal functors.)

Thanks in advance.


1 Answer 1


When you work with lax, oplax or strong monoidal functors, you want to work with bicategorical adjunctions (there will be no $1$-categorical adjoint for sure). The bijection on the Hom sets is thus replaced by an equivalence of categories.

For example, the bicategorical left adjoint of the forgetful functor $\mathsf{MonCat}_{\mathsf{strong}} \to \mathsf{Cat}$ maps a category $\mathcal{C}$ to the monoidal category $M(\mathcal{C})$ of sequences $(X_1,\dotsc,X_n)$ of objects in $\mathcal{C}$ with the obvious (in fact, strict) monoidal structure. If $(\mathcal{D},\otimes)$ is a monoidal category, then the restriction map $$\mathrm{StrongMonFun}(M(\mathcal{C}),(\mathcal{D},\otimes)) \longrightarrow \mathrm{Fun}(\mathcal{C},\mathcal{D})$$ is an equivalence of categories, not an isomorphism of categories (or sets).

It turns out there is no bicategorical left adjoint for $\mathsf{MonCat}_{\mathsf{lax}}$. Otherwise, the image of the left adjoint at the empty category would yield a bicategorical initial object in $\mathsf{MonCat}_{\mathsf{lax}}$. This is a monoidal category $\mathcal{M}$ such that for every monoidal category $\mathcal{C}$ there is an equivalence of categories $$\mathrm{LaxMonFun}(\mathcal{M},\mathcal{C}) \simeq \star.$$ In other words, there is a lax monoidal functor $\mathcal{M} \to \mathcal{C}$, every other is isomorphic to it, and it has only trivial monoidal endomorphisms.

But now consider the special case $\mathcal{C} = \mathcal{M}$ and the identity $\mathrm{id} : \mathcal{M} \to \mathcal{M}$ as well as the trivial lax monoidal functor $1 : \mathcal{M} \to \mathcal{M}$ which, as a functor, is the constant functor at the unit object $1$, and the lax monoidal structure consists of the evident morphisms $1 \to 1$, $1 \otimes 1 \to 1$. By assumption, we have $\mathrm{id} \cong 1$. By unpacking the definition of an isomorphism of lax monoidal functors, this implies that $\mathcal{M}$ is (as a monoidal category) equivalent to the terminal monoidal category $\{1\}$ which has just one object, one morphism, and the trivial strict monoidal structure.

However, we have an isomorphism of categories $$\mathrm{LaxMonFun}(\{1\},\mathcal{C}) \cong \mathrm{Mon}(\mathcal{C}),$$ where $\mathrm{Mon}(\mathcal{C})$ denotes the category of monoids internal to $\mathcal{C}$. So we get $$\mathrm{Mon}(\mathcal{C}) \simeq \star$$ for all monoidal categories $\mathcal{C}$, which is clearly absurd.

For $\mathsf{MonCat}_{\mathsf{oplax}}$ a similar argument works. If there was a bicategorical initial object, it must be $ \{1\}$ again, which together with $$\mathrm{OplaxMonFun}(\{1\},\mathcal{C}) \cong \mathrm{CoMon}(\mathcal{C})$$ yields a contradiction.

Since there is no bicategorical initial object, there is a fortiori no initial object in the $1$-categorical sense.


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