# Monoid and group objects in slice categories

What are the abelian group objects in a slice category $\mathsf{Mon}/X$ of monoids, and a slice category $\mathsf{Grp}/X$ of groups?

I think the latter are perhaps split abelian extensions, but I'm not sure how to prove this and so I'm not sure about the case of monoids either.

How to approach unravelling internal structures in slice categories anyway?

Update and clarification. I would like to somewhat update my question. I have now read the following statements:

1. In the slice category $\mathsf{Grp}/G$, the abelian group objects are the split abelian extensions of $G$.
2. In the slice category $\mathsf{Grp}/G$, the abelian objects - those admitting an internal Mal'cev operation - are the abelian extensions of $G$.

I would like to understand how to prove these statements. At the set level I understand the internal structure unpacks fiberwise with some coherence conditions imposed by the homomorphy of group homomorphisms. What I don't understand is how to actually arive at the characterizations involving abelian kernel.

In his answer, Pece suggests the relevance of the isomorphy of all fibers of a group homomorphism (which happens because of inverses), but still I cannot arrive at the desired conclusions.

I think my question about monoids is answered by the unpacking of internal structure - abelian group objects in slices are nothing more than fiberwise monoids with coherence conditions.

• The statement is false: the trivial $1 \to G$ has abelian kernel but is certainly not a group object in the slice, as it would require it to have a section. – Pece Nov 30 '16 at 14:07
• @Pece ah, I realize what I have overlooked - abelian objects in Mal'cev category need not coincide with abelian group objects, since a Mal'cev category need not be unital. I will edit my mistake out of the question. Thanks! – Arrow Nov 30 '16 at 14:23
• @Arrow: The big idea that is missing from the current answer is to use the fact that all of the structure maps are homomorphisms of monoids rather than just functions of sets. You have two multiplication operations at play, one coming from the fact all objects are monoids, and one coming from the structure of being abelian group object, and you can get identities relating them. – Hurkyl Dec 9 '16 at 12:23
• @Hurkyl yes this is precisely what I can't get out of - using these identities to obtain the characterizations in my question. For instance, if $m$ is an internal binary operation on a monoid homomorphism $f:A\to B$ and $m_b$ is the induced operation on the fiber of $b$, then we have $m_{b_1b_2}(a_1a_2,a_1^\prime a_2^\prime )=m_{b_1}(a_1,a_1^\prime)m_{b_2}(a_2,a_2^\prime )$, but I just don't see how to make use of this. – Arrow Dec 9 '16 at 12:33

Ok, let's unravel what is an abelian group in $\mathsf{Mon}/X$.

Structure. Well it is certainly an object $f : M \to X$ together with a unit morphism $u : 1 \to f$, a multiplication $m : f \times f \to f$ and an inverse $i : f \to f$. But all that is happening in the cartesian category $\mathsf{Mon}/X$, whose morphisms are commutative triangles over $X$, whose unit $1$ is the identity of $X$ and whose product is the pullback over $X$.

Hence the unit is actually a maps $u : X \to M$ such that $fu={\rm id}_X$, meaning $u$ is a section of $f$. The multiplication is then $m : M \times_X M \to M$ satisfying that for $x,y \in M$ such that $f(x) = f(y)$ then this is also the value of $f(m(x,y))$. Endly, the inverse is a map $i : M \to M$ with the property that $f(x) = f(i(x))$ for every $x\in M$.

That's it for the structure. Let's move on to the axioms (i.e. the commutatiity of certain diagrams).

Axioms. One have to interprete the associativity axiom which says $m\circ (m\times {\rm id}) = m\circ ({\rm id} \times m)$. In algebraic terms, here it says: for all $x,y,z \in M$ such that $f(x)=f(y)=f(z)$, $$m(x,m(y,z)) = m(m(x,y),z)$$

The multiplication by unit axiom usually says $m\circ (u\times {\rm id}) = {\rm id} = m\circ ({\rm id} \times u)$. In here, it gives: for all $x \in M$, $$m(uf(x),x) = x = m(x,uf(x))$$

Next is the multiplication by inverse axiom : $m\circ (i\times{\rm id}) = {!}\circ u = m\circ ({\rm id}\times i)$ where $!$ denotes the unique map to the cartesian unit. So here, we get: for all $x \in M$, $$m(i(x),x) = uf(x) = m(x, i(x))$$

Finally, the abelianity axiom, which comes at $m = m \circ \sigma$ with $\sigma$ the canonical braiding of the cartesian structure, gives us: for all $x,y \in M$ such that $f(x) = f(y)$, $$m(x,y)=m(y,x)$$

And we're done! Well, it does seem we can try to say more.

Interpretations. If you look carefully, the axioms always talk about elements in the same fiber of $f$. And those curiously look like the usual axioms of abelian group theory after that.

Lemma. A monoid morphism $f : M \to X$ is an abelian group object (naming $u,m,i$ the structure) in $\mathsf{Mon}/X$ if and only if each fiber $f^{-1}(x)$ is an abelian group (in $\mathbf{Set}$) with unit $u(x)$, multiplication the restriction of $m$ and inverse the restriction of $i$.

So an abelian group object in $\mathsf{Mon}/X$ is more or less a family of abelian groups indexed by $X$ with a coherent choice of unit, multiplication and inverse. (You probably could make that formal, I don't have the patience to do so.)

You can carry the same reasoning in any category $\mathsf A$ instead of $\mathsf{Mon}$ as long as there is forgetful functor $\mathsf A \to \mathsf{Set}$ that preserves pullbacks. So in particular in $\mathsf{Grp}$ : you'll see that not much changes, but in the case of $\mathsf{Grp}$ the interpretation can be carried out further away as each fiber is iso to the kernel...

Remark. We could probably drop the "abelian" part in $\mathsf{Mon}$ and $\mathsf{Grp}$ without changing the outcome because of the Eckmann-Hilton argument.

• Dear Pece, I ended up working out your answer for myself, so I see that at the set level we have fiberwise structure. The structure of arrows then relates the structures on the fibers via coherence conditions, as you wrote (perhaps these need to be added to the Lemma?). However, I still don't see how to deduce the Kernel of an abelian group object in a slice category groups is abelian. I'm sure the key is the isomorphy of all fibers, as you write, but I'm at a loss :) – Arrow Dec 1 '16 at 17:15
• This answer seems lacking in that it doesn't even mention trying to take advantage of the fact that $m,u,i,f$ are all monoid homomorphisms rather than just functions on sets. – Hurkyl Dec 9 '16 at 12:21

There is a well known equivalence due to Jon Beck in his thesis (http://www.tac.mta.ca/tac/reprints/articles/2/tr2.pdf, see discussion starting on page 33), between group objects in slice categories of algebras such as $Grps/G$ and the category of (Beck) modules, in this case just $G$-modules. This was proved by Beck and then used by Quillen in his development of cohomology of commutative algebras. There is an Abelianisation functor from $Grps/G$ to $G-mod$ which, to a group homomorphism, $f:H\to G$ assigns the module of $f$-derivations in the sense of Crowell. (This links up with Fox derivatives in Knot theory and the theory of group presentations and plays a role in more recent work on analogies between knots and prime numbers.)

For the monoid case similar results hold, but are much harder to find in the literature. They are related to the cohomology of small categories due to Wells (http://case.edu/artsci/math/wells/pub/pdf/catext.pdf) who extended ideas of Leech for monoids, for which see the references of the paper by Wells. The connection is not made explicit but is not that difficult to work out. There are links with Baues-Wirsching cohomology as well.

Look at https://ncatlab.org/nlab/show/Beck+module for a summary and some links.