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I've been thinking of a group as a category with one object such that all the arrows are invertible. I have a few questions:

First, am I correct in thinking that a natural transformation between group homomorphisms $f,g: A \rightarrow B$ would be elements $b\in B$ such that $bf = gb$?

This next question may be stupid. I know that homomorphisms induce continuous maps of the classifying spaces, and natural transformations induce homotopies between those maps. Since the classifying space $\mathbb{Z}$ has a the homotopy type of a circle, do homomorphisms $f:\mathbb{Z} \rightarrow G$ give us maps that are essentially paths? And can we define a product between such homomorphisms that agrees with the product in the fundamental group?

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If by $bf$ you mean $f$ followed by the homomorphism corresponding to $b$ (e.g., left multiplication by $b$), and similarly for $gb$, then the answer to your first question is "yes"; I'd need to think about the second question... – Arturo Magidin Aug 11 '11 at 18:25
If you think of homomorphisms $f:\mathbb{Z} \rightarrow \mathbb{Z}$, you get maps from a distinct homotopy class for every integer. It so happens that the fundamental group of a circle is $\mathbb{Z}$. – Joe Aug 11 '11 at 18:36
@Arturo: left multiplication does not yield a group homomorphism, and even if you took conjugation instead, $gb$ wouldn't be defined because $g$ has domain $A$, not $B$. What we can say is that if $b$ defines a natural transformation from $f$ to $g$, then $g=c_b \circ f$ where $c_b$ is the inner automorphism of $B$ induced by conjugation by $b$. – Aaron Aug 11 '11 at 21:42
@Aaron: The arrows when you view a group as a category are not group homomorphisms: they correspond to elements. When you view a group $G$ as a category, the category has a single object (which you may call $G$), and for each $g\in G$, you have an arrow $G\to G$ called "g"; composition of arrows corresponds to multiplication of elements, inverse arrows to inverse elements, and the identity arrow to the identity element of $G$. These "morphisms", which are morphisms in the category sense, are not supposed to be group homomorphisms. – Arturo Magidin Aug 12 '11 at 3:24
@Arturo: But a functor between two groups viewed as categories in this way is a homomorphism: $F(1)=1$ and $F(ab)=F(a)F(b)$. In the question, $f,g$ are functors (homomorphisms) from $A$ to $B$, but $b$ is a group element. As written, $bf=gb$ looks like composition of arrows, but it isn't, because $f$ and $g$ are homomorphisms while $b$ is a group element (a morphism in the category $B$). You specifically say the "homomorphism corresponding to b (e.g. left multiplication by b)." Still, You cannot multiply by $b$ and then apply $g$. See my answer below for an unraveling of the definitions. – Aaron Aug 12 '11 at 3:39
up vote 6 down vote accepted

I don't know much about classifying spaces, but as for the last question, $\mathbb{Z}$ is the free group on one generator; consequently, $\text{Hom}(\mathbb{Z}, G)$ is canonically isomorphic to $G$. This is more than a set, though; it has a group structure which is natural in $G$, so by the Yoneda lemma it follows that $\mathbb{Z}$ has a cogroup structure inducing the group structure on $G$. The cogroup operation $\mathbb{Z} \to \mathbb{Z} \ast \mathbb{Z}$ (where $\ast$ denotes the free product) sends a generator of $\mathbb{Z}$ to the product of the two generators of the two copies of $\mathbb{Z}$, and you can verify that it induces the usual group structure on $G$. If you get stuck on this, see this blog post.

(Note that the pointwise product of homomorphisms is not a homomorphism in general unless $G$ is abelian.)

(Note also that $S^1$ is a cogroup in the pointed homotopy category, and the fundamental group functor sends this cogroup structure to that of $\mathbb{Z}$.)

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That's great information, and it's blowing my mind that this works! Thanks a lot. – Joe Aug 11 '11 at 18:51

For (1), it's worth doing the calculation explicitly, as it is the simplest possible situation in which to have a natural transformation, and thus useful for building intuition. A natural transformation $\eta:F\Rightarrow G$ between two functors $F,G:\mathcal C \to \mathcal D$ is a collection of morphisms $\eta_c\in \mathcal D(F(c),G(c))$ for each $c\in \mathcal C$ such that, then $\eta_{c'}F(f)=G(f)\eta_c$ for every $f:c\to c'$

When $\mathcal{C}, \mathcal{D}$ have one object, $\eta$ is described by a single morphism in $\mathcal{D}$ which intertwines between $F$ and $G$. Switching notation back to the variables in the original equation, if $A$ and $B$ are groups and $f$ and $g$ are homomorphisms, then a natural transformation from $f$ to $g$ is a $b\in B$ such that $bf(a)=g(a)b$ for all $a\in A$.

You could shorten this to $bf=gb$, but to me, it isn't immediately clear what that would mean.

Because $B$ is a group, we can rephrase this slightly by writing $bf(a)b^{-1}=g(a)$, or $g=c_b \circ f$ where $c_b$ denotes the inner automorphism of $B$ coming from conjugation by $b$. In this case, every natural transformation is a homotopy equivalence, and the equivalence classes of are homomorphisms up to conjugation. I believe that there are connections between this and what you have asked in (2), but I do not currently have a complete answer to give.

If $f=g$, so that we can compose two natural transformations, we see that $\operatorname{End}(f)$ is the centralizer of $f(A)$ in $B$, that is, the elements of $b$ which commute with $f(a)$ for all $a\in A$. If we specialize further to the case that $A=B$ and $f=g=\operatorname{id}_A$, we have $\operatorname{End}(\operatorname{id}_A)=Z(A)$, the center of $A$. This generalizes slightly: if $A$ is an abelian category with one object (which is equivalent to being a ring), the natural transformations between the identity functor has a ring structure and will be the center of $A$ viewed as a ring. More generally, we can define the Bernstein center $Z(\mathcal C):=\operatorname{End}(\operatorname{id}_{\mathcal C})$, which for regular categories will be a monoid and which for enriched categories will have additional structure (I believe that it will be a monoid in the category you are enriching over).

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(Re: 1) Yes, definitely (so, for example, endofunctor equivalent to identity = inner automorphism).

(Re: 2) Sure. $\operatorname{Hom}(\mathbb Z;G)=G$ (isomorphism is just $f\mapsto f(1)$). And isomorphism $\pi_1(BG)=G$ is a group isomorphism.

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