What are the subcategories of ordered sets / groups?

Question

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.)

Answer 1

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.

Answer 2

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 $Z(G{-}\mathbf{Set}) \cong Z(G)$.