Given a 2-group $\mathcal{G}$, you can construct a crossed module $(G,H,t,\alpha)$ and vice versa.

Is there something similar you can say for strict 2-categories?

In a personal attempt to understand strict 2-categories, I ended up constructing a speculative conceptual tool (whose validity remains to be seen) that I call the boundary of a 2-morphism. I've written up some raw notes here:

The basic idea is that given morphisms $f,g:x\to y$ and a 2-morphism $\alpha:f\Rightarrow g$, we define its boundary as an endomorphism

$$\partial\alpha:y\to y$$


$$\partial\alpha\circ f = g.$$

When the source of the 2-morphism is an identity morphism, then we have

$$\partial\alpha = t(\alpha),$$

which seems to relate things well to cross modules when all morphisms are invertible.

I'm curious if there is anything like a crossed module, but where we're not dealing with groups and morphisms are not invertible. What I'm trying to cook up seems like it might be related to such a thing if it exists.

Any thoughts and/or any comments on my notes would be greatly appreciated.

PS: Apologies in advance if my writing is not very clear. I'm not a mathematician, but am trying to teach myself some basic higher category theory.

  • $\begingroup$ It might make sense to first ask your question for 2-categories with one single object. Indeed, 2-groups are 2-categories with one single object, and with the additional requirement that all 1- and 2-morphisms are invertible. $\endgroup$ Aug 15, 2010 at 6:52

2 Answers 2


This is not an answer, but rather a "no go" observation. I claim that you should not expect 2-categories in general to have "crossed-module" like descriptions, or at least not any such description that's any easier to think about than "2-category". Part of what makes 2-groups easy is that they have lots and lots of symmetry. Ignoring the 2-morphisms (and 2-composition), the 1-morphisms form a group, so by group translation you can relate the structure between any two 1-morphisms to the structure between some 1-morphism and the identity. And that structure is group- or torsor-like, since if you ignore the 0-morphism and the 1-composision, the 1-morphisms are a groupoid.

I expect that you can construct something for a 2-category with (1) only one 0-morphism and (2) all 1-morphisms invertible. I.e. this is a 2-group but relaxing the invertibility condition on arbitrary 2-morphisms. Then I would expect that this should correspond to a "crossed module of groups" where the second "group" $H$ need only be a monoid, although I haven't thought about the details.

  • $\begingroup$ Thank you Theo. I'm sure you're right about the "no go" in general. In trying to answer my own question, I had the idea to look at Cat with (small) categories, functors, and natural transformations. If there were no "no go", then every natural transformation would have a boundary. I'm still trying to work out the conditions under which a boundary exists. I don't think the morphisms need to be all invertible, but maybe they need to have a "right inverse" (?) There may be something in between general strict 2-category and 2-groups for which we can define some crossed module-like construction. $\endgroup$
    – EricForgy
    Aug 16, 2010 at 5:57
  • 1
    $\begingroup$ There are various notions in the literature that correspond to analogues of crossed modules where the top group H is just a monoid, usually an algebra. Plenty of these appear for instance in the literature on orbifold/equivariant CFT. These indeed define 2-categories with invertible 1-morphisms and non-invertible 2-morphisms. Maybe I find the time to dig out some references from when I looked into this... $\endgroup$ Aug 16, 2010 at 14:58
  • $\begingroup$ Thanks Urs. I think Theo and Chris have made the "no go" observation clear for general strict 2-categories. But it is interesting to hear about this crossed module-like construction where the top group is just a monoid. I can't help but think there is a little bit further we can go. Perhaps where even the bottom group is a monoid, but this may put restrictions on the allowed 2-categories. For example, I'm thinking perhaps the 2-category must be "directed" somehow. $\endgroup$
    – EricForgy
    Aug 17, 2010 at 5:31

I think Theo's "no go" is exactly right. Here is an example which might make things easier to understand: Let X be any category. I am going to construct an interesting 2-category with one object which is like a 2-group, but without the invertibility. So there is a single 0-morphism p. The morphisms from p to itself form a category which is a disjoint union of X and two points: $$ 0 \sqcup X \sqcup \infty $$ This is the disjoint union of categories so X and these other points don't interact. That completely describes the vertical composition. Now I need to tell you the horizontal composition. The element 0 is the (strict) identity for the horizontal composition. The point $\infty$ has the property that $z \cdot \infty = \infty = \infty \cdot z$ for any z. Finally the horizontal composite of any two things in X results in $\infty$.

Equivalently we can describe this as a monoidal structure on $ 0 \sqcup X \sqcup \infty $. It is actually strictly commutative too.

The reason this an important example is that we have embedded the category X fully-faithfully into this monoidal category. So any sort of algebraic description of monoidal categories or 2-categories or even strict 2-categories must be at least as complicated as the theory of all categories. This is in severe contrast with the situation for 2-groups for the reasons that Theo pointed out.

This example is also related to Reid Barton's answer to my question: Hom alg for comm. monoids. See also the related questions: A peculiar model strcture on simplicial sets? and simplicial commutative monoids group completion. The example I just described also works to give a simplicial commutative monoid where now X is any simplicial set. However when you apply the "Dold-Kan correspondence" you always get the zero chain complex. This shows that the Dold-Kan correspondence fails to be an equivalence for commutative monoids. It also says that in order to describe higher categories in terms of something like a chain complex (e.g. something like a crossed module) you absolutely need some invertablity.

  • $\begingroup$ Awesome example. $\endgroup$ Aug 16, 2010 at 21:45
  • $\begingroup$ I agree with the answer that crossed modules require the invertibility. However broadening rather than narrowing does work! That is going from globular or simplicial to cubical; see the paper Al-Agl/Brown/Steiner `Multiple categories: the equivalence between a globular and cubical approach', Advances in Mathematics, 170 (2002) 71-118. $\endgroup$ Feb 7, 2016 at 12:30

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.