## How strict can I be in the definition of “2-group”?

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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.

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## 2 Answers

You can always do this. Take any $b$ and define $d = p_2 b$. Then $b' = (p_1, d)$ is equivalent to the original $b$. To see this note that

$$(p_1, m) \circ b = (p_1b, m \circ (p_1 b, d)) \simeq id = (p_1, p_2)$$

The first component shows $p_1 b \simeq p_1$. We use this transformation $\times id$ to show that $b \simeq b' = (p_1, d)$.

Now we consider the equivalence $(p_1, m) \circ b' \simeq id$. Here we have,

$$(p_1, m) \circ (p_1, d) = (p_1, m \circ (p_1, d)) \simeq id = (p_1, p_2)$$

restricting to $G = G \times { 1} \subseteq G \times G$, this gives a natural isomorphism $m(x, d(x, 1)) \simeq 1$. You can take $i(x) = d(x,1)$, and we have $x i(x) \cong 1$.

We also have

$$(p_1, d) \circ (p_1, m) = (p_1, d \circ (p_1, m)) \simeq (p_1, p_2)$$

which gives a natural isomorphism, $d(x, xy) \simeq y$ (writing $m(x,y) = xy$). Thus we have,

$$d(x,y) \simeq d(x,1 y) \simeq d(x, x i(x) y) \simeq i(x) y,$$

which is the formula you were after. So we can replace $d(x,y)$ with $m(i(x), y)$ to get a third inverse functor b''. Note that this doesn't mean that we have a strict 2-group, just that we can define the inverse functors and difference functors you asked about.

Notice also that we didn't really use anything about G being a 1-category as opposed to an n-category (except the associator and unitors) so this argument generalizes to the n-group setting basically verbatim.

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As an additional remark, it is also possible to choose isomorphisms $i(x) \cdot x \cong 1$ in such a way as to be compatible with the isomorphism $x \cdot i(x) \cong 1$. This takes a little more work, but the argument is explained in the paper by Baez and Lauda on 2-groups.  – Chris Schommer-Pries Feb 16 2010 at 21:28
The Baez-Lauda paper is available here: arxiv.org/abs/math/0307200 – Chris Schommer-Pries Feb 17 2010 at 3:47

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I think the answer is no, but cannot give a counterexample off hand. What is true is that there will be an equivalent monoidal category such that the corresponding functor $b$, does have the property. Any 2-group is equivalent to a strict 2-group (which can be assumed to come from a crossed module). What would be an interesting subsidiary question would be can the structure be `deformed' to one which is strict. (Deformed in the sense of a deformation theory.)

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