Standard definition of 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 Strict 2-Group is a group object in the category of categories and the category object in the category of groups. Now in terms of the last two equivalent formulations of Strict 2-Group 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 Standard definition is actually implying that we have an identity element in the morphism set and each morphism has an inverse with respect to the binary operation in the morphism set induced by the associated bifunctor of the Strict Monoidal category.

Also I could not get how we are getting the associativity of the binary oporation in the the morphism set induced by the associated bifunctor of the strict monoidal category.

I felt from the Standard definition we can only say that object set is a group and the morphism set is equipped with a binary operation NOTE ( Binary operations on both object and morphism sets here are induced by the associated bifunctor of the strict monoidal category.) . But existence of inverse , associativity and identity in the morphism set 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?


Here I am NOT ASKING : Whether each morphism is invertible or not?

The issue of invertibility of morphism(in categorical sense) is already discussed 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]). They discussed about invertibility of arrow in categorical sense ( that is whether a group object in Cat( Category of Categories) has a groupoid structure or not? ..)

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”

The link from where I read the standard definition are the following: https://arxiv.org/pdf/math/0307200.pdf , https://en.wikipedia.org/wiki/2-group#Strict_2-groups

If my question is stupid or not up to the standard of this forum then I am seeking apology before hand .

Thank you.

  • 2
    $\begingroup$ I thought the same about the MO version of your question, but could you please use less formatting ? Right now it's just confusing and distracting. $\endgroup$
    – Arnaud D.
    Commented Jun 26, 2019 at 6:44
  • $\begingroup$ @ArnaudD. Yes it is basically the same question. I did not get answer there(MO) even after specifying the reason that my question is not duplicate to the question you stated. So I asked it here again to get some helpful answer. $\endgroup$ Commented Jun 26, 2019 at 6:47
  • $\begingroup$ @ArnaudD. I used more formatting because I do not want others to get confused with my question (like the last time you got confused and said my question is duplicate to that question). In fact invertibility of morphism was never my question.(Which I mentioned here explicitly) $\endgroup$ Commented Jun 26, 2019 at 6:51

1 Answer 1


The associativity and existence of an identity on the set of morphism is already given by the strict monoidal structure. Indeed, the functors $(A,B,C)\mapsto (A \otimes B) \otimes C$ and $(A,B,C)\mapsto A\otimes (B\otimes C)$ are equal, hence also their morphism parts are equal; and similarly, the functor $A\mapsto I\otimes A$ is equal to the identity functor, thus also at the level of morphism, where it is defined by $f\mapsto 1_I\otimes f$, which shows that $1_I$ is an identity for the set of . Note that this part is true for any strict $2$-category.

Inverses are a little bit more tricky : here we need to use some sort of variant of the Eckmann-Hilton argument. More specifically, note first that if $A\otimes \widehat{A}=I$ then we have $1_A\otimes 1_{\widehat{A}}=1_{A\otimes \widehat{A}}=1_I$, so identity maps have inverse for $\otimes$.

Moreover for any $f:A\to B$ and $g:C\to D$, we have, by the functoriality of the tensor $\otimes$ : $$f\otimes g=(1_B\circ f)\otimes (g\circ 1_C)=(1_B\otimes g)\circ (f\otimes 1_C).$$ In particular, if we want $\widehat{f}:\widehat{A}\to \widehat{B}$ to be the inverse of $f$ for the operation $\otimes$, we must have $$1_B\otimes \widehat{f}=(f\otimes 1_{\widehat{A}})^{-1}=f^{-1}\otimes 1_{\widehat{A}}^{-1}=f^{-1}\otimes 1_{\widehat{A}},$$ and multiplying both sides by $1_{\widehat{B}}$ shows that we must choose $\widehat{f}=1_{\widehat{B}}\otimes f^{-1}\otimes 1_{\widehat{A}}$. Now we can check that $\widehat{f}\otimes f =1_I$ using the interchange law as above.

  • $\begingroup$ Thanks! I am trying to understand your answer. $\endgroup$ Commented Jun 26, 2019 at 11:07
  • $\begingroup$ Thanks a lot! I now got my answer. $\endgroup$ Commented Jun 26, 2019 at 12:41

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