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We define a category $C$ to be $Grp$-Enriched if for every $X$,$Y\in C$, we have that $C(X,Y)$ 'has group structure'/is a group. But what does that mean really? If I am given any finite set $A$, there is a group of order $|A|$, $Z_n$, so may I say "A has a group structure"? But I imagine that's not what they mean when they say the hom sets have group structure. But then 'where' is this structure? Is there a canonical way this structure is imposed across all hom sets? Because if $C(X,Y)$ is just a set of morphisms, and we say we can look at it as a group, isn't that true of every set?

I feel like I'm just looking at it in the wrong way. I suppose a related question (which I think might underscore my confusion better), when we say $G$ is an object in the category $Grp$, and thus $G$ is a group. How are we looking at $G$? Are we looking at $G$ to be a set G with a product map $\bullet$? Or am I stuck in set theoretic ways of thinking?

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    $\begingroup$ I'm not familiar enough with enrichment to offer a good answer, but an object of Grp is defined as a pair $(G,\cdot)$ satisfying the axioms (...or possibly some equivalent definition; it doesn't matter). A category enriched over $\mathbf{M}$ includes, as part of its definition, an object $C(X,Y)$ in $\mathbf{M}$ for each pair of objects - so the data of how each "hom set" is a group is baked into the structure. (...and then there are some axioms for how these objects relate and a bit more data included in the structure) $\endgroup$ – Milo Brandt Dec 11 '19 at 16:10
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To talk about enrichment in $\mathcal{Grp}$, we need a bit of structure on it first. The traditional structure is that of a monoidal category. The only two monoidal structures on $\mathcal{Grp}$ that I'm aware of are those given by the cartesian product and the free product (the product and coproduct respectively). The unit for both of these is the trivial group. Abelian groups have a different monoidal structure given by the tensor product with $\mathbb{Z}$ as the unit.

The first bit of structure in an enriched category is a set/collection of objects. Then, for every pair of objects $X$ and $Y$, we have a group $C(X, Y)$. A priori, this doesn't even need to come from the hom set of some category (though the rest of the structure below suffices to prove that this is the case). You could potentially get a category enriched in $\mathcal{Grp}$ by starting with a single object $\bullet$ and defining $C(\bullet, \bullet)$ to be $\mathbb{Z}$

But besides each hom being a group, the rest of the category structure has to respect the group structure. Instead of an identity morphism $id_X \in C(X, X)$ for each object $X$, we need a group homomorphism $id_X: I \to C(X, X)$ for each object $X$ (where $I$ is the unit for the monoidal structure on $\mathcal{Grp}$, i.e. the trivial group if we use the cartesian product as the monoidal product). The composition map $m: C(Y, Z) \otimes C(X, Y) \to C(X, Z)$ also needs to be a group homomorphism.

Of course, besides all this, these maps have to be natural in an appropriate way (extranatural), composition has to be associative, the identities have to really be identities, and so on. See enriched category for more information.

I'll let you work out the details of what that means for group-enriched categories, but I hope I've shown that you need a lot more than the mere structure of a group on every hom set.

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It is probably best to have some examples of categories enriched in groups. The categories of abelian groups and of vector spaces over some field are the examples one meets earliest on: one has pointwise operations giving each Hom-set a natural group structure. The natural aspect of enrichment is very important: one can’t simply choose arbitrary group structures, because the composition will generally not respect those choices.

As described in another answer, to say that a category $C$ is enriched in $C$ means that we have Hom-objects $C(x,y)$ in $V$ for objects $x,y$ in $C$ and a composition morphism $C(x,y)\otimes C(y,z)\to C(x,z)$-which is, again, a morphism in $V$. One has to pick a monoidal product $\otimes$, which is roughly speaking a monoidal structure on the category $V$. For the case of groups, the most natural monoidal structure would just be the Cartesian product, so we would need that the composition be a homomorphism from the product of Hom-groups, that is, $f*g\circ h*k=f\circ h*g\circ k$ if $f,g:y\to z$ and $h,k:x\to y$, where $*$ denotes the group structure and $\circ$ the composition.

That said, this property holds in no standard examples: categories enriched in groups, in this sense, effectively do not exist. In particular, the category of groups is not enriched over itself! Instead, the groups in the standard examples are abelian, and the composition is bilinear, so that these categories are enriched over abelian groups endowed with the tensor product as a monoidal structure. Note that the category of abelian groups is naturally enriched over itself in this sense; such closed monoidal categories are generally the only ones one enriches over.

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