Center of the categories $\mathbf{Grp}$ and $\mathbf{Ab}$.

Question

This is Exercise II.5.8 from Mac Lane, Categories for the Working Mathematician.

For the identity functor $I_C$ of any category, the natural transformations $\alpha:I_C\dot{\to}I_C$ form a commutative monoid. Find this monoid in the cases $C=\mathbf{Grp}$, $\mathbf{Ab}$ and $\mathbf{Set}$.

I found out that this monoid is called the center of the category $C$, and I'm going to call is $Z(C)$. My problem was to find a nice characterization of the centers of $\mathbf{Grp}$ and $\mathbf{Ab}$. Here's my work so far:

Notice that $\alpha:I_C\dot{\to} I_C$ is natural iff $\alpha_c$ commutes with every arrows $f:c\to c$ for every $c$ (in other words, iff $\alpha_c$ is in the center of the monoid $\hom_C(c,c)$). In symbols: $$Z(C)=\left\{\alpha:I_C\dot{\to}I_C \mid \forall c\in C, \forall f\in\hom_C(c,c), \alpha_c f=f\alpha_c\right\}\qquad (*)$$ This implies directly that the center of $\mathbf{Set}$ is trivial (that is, it contains only the unit natural transformation).

However, this will not work in $\mathbf{Grp}$ and $\mathbf{Ab}$, since the monoid of endomorphisms of $\mathbb{Z}$ is commutative. However, not every two endomorphisms of arbitrary groups commute: Let $G=F_2$, the free group with two generators $a$ and $b$. Let $f,g:F_2\to F_2$ be morphisms such that $f(a)=f(b)=ab$, $g(a)=g(b)=a$. Then $fg(a)=f(a)=ab$, and $gf(a)=g(ab)=a^2$.

That said, the best description of $Z(C)$, for $C=\mathbf{Grp}$ or $\mathbf{Ab}$, I can give at the moment is ($*$) as above. What would be better descriptions of $Z(\mathbf{Grp})$ and $Z(\mathbf{Ab})$?

PS: Since I'm self-studying, I wouldn't tag this question as "homework".

Answer 1

The key idea you're missing is to use generators of your category.

In $\mathbf{Set}$, the terminal element generates the category; every map $A \to B$ is completely determined by its composites with maps $\mathbf{1} \to A$ which are a kind of "generalized element", and the fact $\alpha$ is natural lets us understand $\alpha_A$ in terms of $\alpha_\mathbf{1}$.

Writing $a$ for the map $\mathbf{1} \to A$ that sends the unique element of $\mathbf{1}$ to $a \in A$, we have

$$ \begin{matrix} \mathbf{1} & \xrightarrow{\alpha_1} & \mathbf{1} \\ \ \ \downarrow a & & \ \ \downarrow a \\ A & \xrightarrow{\alpha_A} & A \end{matrix} $$

which implies that $\alpha_A(a) = a$ for all $a \in A$.

The categories $\mathbf{Grp}$ and $\mathbf{Ab}$ are both generated by $\mathbb{Z}$, and the same diagram tells us what we need to know:

$$ \begin{matrix} \mathbb{Z} & \xrightarrow{\alpha_\mathbb{Z}} & \mathbb{Z} \\ \ \ \downarrow x & & \ \ \downarrow x \\ G & \xrightarrow{\alpha_G} & G \end{matrix} $$

where this time I've written $x$ for the map that sends $1$ to $x \in G$. This tells us

$$ \alpha_G(x) = x^{ \alpha_\mathbb{Z}(1)}$$

so all we need to do is to check which endomorphisms of $\mathbb{Z}$ yield natural transformations when extended to the whole category in this way.

For Abelian groups, for each integer $n$ the operation $x \to x^n$ is indeed a natural transformation, so we see that Z(Ab) is the multiplicative monoid of all integers. (in fact, if you consider the additive structure, it is the ring of integers)

For general groups, $x \to x^n$ generally isn't even a group homomorphism. For example, the $n=2$ case would ask for $(xy)^2 = x^2 y^2$, and the $n=-1$ case would ask for $(xy)^{-1} = x^{-1} y^{-1}$, both of which are equivalent to the condition that $x$ and $y$ commute.

Only the cases of $n \in \{ 0, 1 \}$ give group homomorphisms for every group, and both are easily seen to be natural transformations. Thus Z(Grp) is this two element multiplicative monoid.