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.
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.