Recall that a *group* is an associative, unital monoid $G$ such that the map $(p_1,m) : G \times G \to G\times G$ is an isomorphism of sets. Here $p_1$ is the first projection and $m$ is the multiplication, so the map is $(g_1,g_2) \mapsto (g_1,g_1g_2)$.

My question is a basic one concerning the definition of "2-group".

Recall that a *monoidal category* is a category $\mathcal G$ along with a functors $m : \mathcal G \times \mathcal G \to \mathcal G$ and $e: 1 \to \mathcal G$, where $1$ is the category with one object and only identity morphisms, such that certain diagrams commute up to natural isomorphism and those natural isomorphisms satisfy some axioms of their own (the natural isomorphisms are part of the data of the monoidal category).

Then a *2-group* is a monoidal category $\mathcal G$ such that the functor $(p_1,m): \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ is an equivalence of categories. I.e. there exists a functor $b: \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ such that $b\circ (p_1,m)$ and $(p_1,m) \circ b$ are naturally isomorphic to the identity. Note that $b$ is determined only up to natural isomorphism of functors.

Question:Can I necessarily find such a functor $b$ of the form $b = (p_1,d)$, where $d : \mathcal G \times \mathcal G \to \mathcal G $ is some functor (called $d$ for "division")? If so, can I necessarily find $d = m\circ(i \times \text{id})$, where $i: \mathcal G \to \mathcal G$ is some functor (called $i$ for "inverse")?

In any case, the natural follow-up question is to ask all these at the level of 3-groups, etc.