The standard definition of a strict 2-group says that it is a strict monoidal category in which every morphism is invertible and each object has a strict inverse.

Also it is a well known fact that a strict 2-group is a group object in the category of categories, and also a category object in the category of groups.

Now in terms of the latter two (equivalent) formulations of strict 2-groups, we must have a group structure on both the object set and the morphism set of the strict 2-group induced by the associated bifunctor of the strict monoidal category. But I am not able to understand how the standard definition actually implies that each morphism has an inverse with respect to this binary operation. Also I could not get how we are getting the associativity of this binary operation on the the morphism set.

I felt that from the standard definition we can only say that the object set is a group and that the morphism set is equipped with a binary operation with respect to which it has an identity element. (NOTE: we are talking about the binary operations on both object and morphism sets induced by the associated bifunctor of the strict monoidal category.) But, it seemed to me that existence of inverses and associativity are not guaranted.

Hence if those two equivalent formulations are true, and also the standard definition is correct, then I must be misunderstanding something here conceptually.

Then where am I making the mistake?

If my question is stupid or not up to the standard of this forum then I apologize beforehand.

This is not a duplicate of the question strict 2-groups VS crossed modules, because I felt in that question they discussed the invertibility of arrows in the categorical sense (that is, whether a group object in the category of categories has a groupoid structure or not); this fact is trivial and also mentioned in the paper by R. Brown and C. Spencer, G-groupoids, crossed modules and the fundamental groupoid of a topological group, Proc. Kon. Ned. Akad. v. Wet, 79, (1976), 296 – 302, [pdf]. Here I am asking about the existence of group inverses in the morphism set with respect to the binary operation induced from the bifunctor of the strict 2-group (not whether each morphism is invertible or not).

I hope my above explanation is sufficient to address the issue. If not, I will try to give more detail.

Thank you.

  • $\begingroup$ @ArnaudD. Sorry I couldn’t get how my question is duplicate of “ Strict 2-groups vs crossed module “ ..please can you elaborate ? I couldn’t not see much relation. $\endgroup$ Jun 22, 2019 at 0:38
  • $\begingroup$ @ArnaudD. I felt in that question they discussed about invertibility of arrow in categorical sense ( that is whether a group object in Cat has a groupoid structure or not? .. this fact is trivial ( also mentioned in the Brown and Spencer paper) Here I am asking about the existence of group inverse in the morphism set with respect to the binary operation induced from the “ Bifunctor of the Strict 2 - group” $\endgroup$ Jun 22, 2019 at 0:47

1 Answer 1


If I understand the question correctly, then the answer for associativity is that the associator isomorphism in a non-strict monoidal category is natural, and that this naturality condition is still there in a strict monoidal category when the components of the associator are identity maps. Put differently, in a strict monoidal category, the two functors $C\times C\times C\to C$ don't just act equally on objects, they are equal as functors.

The construction of tensor-inverses for morphisms from the (a priori non-functorial) mere existence of tensor-inverses for objects is a bit trickier. Suppose $f:x\to y$, with composition-inverse $g:y\to x$ (so that $f\circ g = 1_y$ and $g\circ f = 1_x$); then the tensor-inverse of $f$ is the morphism

$$ x^{-1} = y^{-1} \otimes y \otimes x^{-1} \xrightarrow{1\otimes g \otimes 1} y^{-1} \otimes x \otimes x^{-1} = y^{-1}. $$

That this is a tensor-inverse of $f$ follows from a calculation involving naturality, most naturally (npi) done with string diagrams where the equalities such as $x \otimes x^{-1} = I$ are indicated by morphisms forming a dual pair (that just happen to be identity morphisms in this case).

  • $\begingroup$ Thanks for the Answer! $\endgroup$ Jun 26, 2019 at 12:03
  • $\begingroup$ Thanks a lot!! I got my answer!! $\endgroup$ Jun 26, 2019 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.