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This is an exercise from Tom Leinster's Basic Category Theory. It asks:

1) What are the subcategories of an ordered set? Which are full?

2) What are the subcategories of a group? Which are full?

I'm self studying and I couldn't discuss this with someone, so I'm curious for what the solution is.

For the second one I think, if you have a one object group category, no subcategories are full, as they all consist of the group object with fewer morphisms. Also all of them are monoids, but only the ones that contain the inverse of every morphism remain groups, with less elements than the original.

Also a related question:

Is there a functor $Z:\mathbf{Grp} \to \mathbf {Grp}$ so that $Z(G)$ is the center of the group $G$?

(I'm guessing there is one, and also that the center of $G$ is a full subcategory of $G$, and also an Abelian group.)

user557
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Mano Plizzi
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2 Answers2

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Interesting questions. Let start be saying that your observations on the subcategories of groups are correct: if you exclude the empty category all the subcategories of a group are exactly the submonoids of the group (by the way there is a subcategory that is full: the group itself).

About the posets, a subcategory of a poset-category $P$ is a poset $Q$ such that as sets $Q \subseteq P$ (that means that $\text{Ob}(Q) \subseteq \text{Ob}(P)$) and for every pair of elements $x,y \in Q$ if $x \leq_Q y$ then $x \leq_P y$.

That doesn't mean that $Q$ is a subposet of $P$ in the classical sense: usually by a subposet of $P$ one would mean a subset $Q \subseteq P$ with the ordering $$x \leq_Q y \iff x \leq_P y\ .$$ These subposets correspond exactly to the full-subcategories of $P$.

Finally about the existance of the functor $Z \colon \mathbf{Grp} \to \mathbf{Grp}$, it will surprise you but the answer is no, such functor does not exists. Here is a proof by contradiction. Assume that such a functor $Z$ does exists, then consider the following group-homomorphisms $$\mathbb Z/2\mathbb Z \stackrel{f}{\longrightarrow} S_3 \stackrel{g}{\longrightarrow} \mathbb Z/2 \mathbb Z$$ where $f(i)=(1,2)^i$ and $g$ is the projection of $S_3$ onto $S_3/A_3\cong \mathbb Z/2\mathbb Z$. You can easily see that $g \circ f=1_{\mathbb Z/2\mathbb Z}$ so we should have $$Z(g)\circ Z(f)=Z(g \circ f)=Z(1_{\mathbb Z/2\mathbb Z})=1_{Z(\mathbb Z/2\mathbb Z)}=1_{\mathbb Z/2\mathbb Z}$$ on the other hand we have the situation shown in the following diagram $$Z(\mathbb Z/2\mathbb Z) \stackrel{Z(f)}{\longrightarrow} Z(S_3)=0 \stackrel{Z(g)}{\longrightarrow} Z(\mathbb Z/2\mathbb Z)$$ so $Z(g)\circ Z(f)$ should be the zero-morphism from $\mathbb Z/2\mathbb Z$ in itself, which is not the identity, hence we arrived to a contradiction. We have to conclude that the functor $Z$ cannot exist.

Hope this help.

Giorgio Mossa
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  • Doesn't the implication $x\le_Q y\implies x\le_P y$ hold for an arbitrary subcollection of objects and arrows in the original category? – user557 Jun 25 '19 at 01:58
  • @user634426 I am not sure that I fully understand your question. Assuming that $Q$ is a category and that its objects and morphisms are objects and morphisms of $P$ (and that the category-structure of $Q$ is the one induced by $P$) then the implication does indeed hold. – Giorgio Mossa Jun 25 '19 at 13:09
  • In your answer, you're saying that a subcategory of a poset category $P$ is a poset $Q$ such that $Q\subset P$ and the implication from my comment holds. But as far as I can see (I must be mistaken), $Q$ need not be a subcategory in order for these two conditions to hold. If $Q$ consists of some collection of objects of $P$ and some collection of arrows of $P$ (without the requirement that the composition of two composable arrows in $P$ is in $P$), then the two conditions hold, but such a $Q$ is not a subcategory of $P$. – user557 Jun 25 '19 at 14:09
  • @user634426use Now I tuink I got it. What you says it is true, but if you don't close by composition $Q$ is not a poset. Composition gives you transitivity, without it you don't have a poset. Hope this solves your doubts. – Giorgio Mossa Jun 25 '19 at 15:14
  • Oh, I see, if we enforce the condition that $Q$ be a subcategory, then the relation in $Q$ that corresponds to the arrows in $Q$ is reflexive and transitive (b/c $Q$ is closed under identities and composition). And that relation is antisymmetric because if there were distinct objects $a,b$ in $Q$ with arrows $a\to b$ and $b\to a$, then the same objects and arrows would belong to $P$, contradicting the fact that $P$ is antisymmetric. So $Q$ must itself be a poset. But what about the empty category? It's also a subcategory (even a full subcategory), right? – user557 Jun 25 '19 at 15:43
  • @user634426 sure. – Giorgio Mossa Jun 25 '19 at 21:03
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Regarding the second question, the center crops up in a different way:

If $G$ is a group viewed as a category and $1_G$ is the identity functor, then $Z(G)$ is the group of natural transformations $1_G \to 1_G$.

It can be useful to extend this to more general categories: for an arbitrary category $\mathbf{C}$, we can define $Z(\mathbf{C}) = \hom(1_\mathbf{C}, 1_\mathbf{C})$.

As a neat example, for any group $G$, if $G{-}\mathbf{Set}$ is the category of left $G$ actions on sets, then $Z(G{-}\mathbf{Set}) \cong Z(G)$.