It has been many years since I first read Categories for the Working Mathematician, but I still have a question about one of the first exercises. Question 5 in section 1.3 asks you to find two different functors $\mathsf{T}: \mathsf{Groups} \to \mathsf{Groups}$ with object function $\mathsf{T}(\mathsf{G}) = \mathsf{G}$ for every group $\mathsf{G}$. I have played with this for a long time, and none of the obvious choices end up working. Was this a mistake on Mac Lane's part, or am I just missing something very obvious?

If it turns out there are no "obvious" choices, does anyone have an idea of how to prove that there are not two such functors?

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    $\begingroup$ The identity functor and the functor sending every morphism to the trivial morphism should also do. $\endgroup$ – HenrikRüping Jan 26 '15 at 11:45
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    $\begingroup$ @HenrikRüping The identity must be sent to the identity. But then, if $G \to H \to G$ is the identity, we can't send the maps $G \to H$ or $H \to G$ to the identity. $\endgroup$ – David E Speyer Jan 26 '15 at 11:56

This is an "evil" question, which deserves an evil answer. Pick your favorite pair of an object G0 of Groups and a nontrivial automorphism φ of G0. Define the functor T : Groups → Groups by T(G) = G, T(f) = [φ ∘] f [∘ φ-1] where we compose with φ if the target of f is G0 and compose with φ-1 if the domain of f is G0. This T is clearly different from the identity functor (it acts by φ on Hom(Z, G0) = the underlying set of G0).

This answer isn't very satisfying though because T is naturally isomorphic to the identity. I don't know whether one can find an example where T is not naturally isomorphic to the identity.

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  • $\begingroup$ Excellent! I thought I had considered this solution (or something very much like it) at some point or another, but it hadn't worked out. I guess I was remembering incorrectly. Thanks! $\endgroup$ – Steven Gubkin Nov 5 '09 at 17:53
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    $\begingroup$ Actually, I think it's quite satisfying. It's important to be aware that the generality of functors and natural transformations still allows them to be fairly "unnatural", and I think this is the simplest example to illustrate that (especially since it works for any object with a non-trivial automorphism in any category). $\endgroup$ – Andrew Critch Nov 5 '09 at 17:55
  • $\begingroup$ Ah, I was thinking that the condition "T(G) = G" was evil, but actually it's the conclusion of whether two functors are equal that's the relevant evil in my example (we should instead ask for an isomorphism in the functor category). $\endgroup$ – Reid Barton Nov 5 '09 at 18:15
  • $\begingroup$ This is a funny question and a funny answer and it would be really funny if there would be a T which is not isomorphic to the identity. If somebody finds out, please post it! $\endgroup$ – user717 Nov 5 '09 at 18:19
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    $\begingroup$ So just to clarify, your functor is trivial on any morphism that does not involve G_0. $\endgroup$ – Theo Johnson-Freyd Nov 5 '09 at 20:41

Since this question popped up again, I might as well sketch how far I got. The trivial morphism can be recognized because it factor through the trivial group, so the trivial morphism must be sent to itself. From this we deduce that the obvious inclusions and projections between $G$, $H$ and $G \times H$ are sent to themselves up to the sort of evil isomorphisms discussed in Reid's answer. For any group $F$ with maps $F \to G$ and $F \to H$, the product map $F \to G \times H$ is sent to the product map, again, up to evil isomorphisms. In particular, the diagonal embedding $G \to G \times G$ is sent to the diagonal up to isomorphisms.

If $G$ is abelian, then $\mu: (g_1, g_2) \mapsto g_1 g_2$ is a map $G \times G \to G$ and analyzing the composition of $\mu$ with the coordinate embeddings $G \to G \times G$ shows that $\mu$ goes to $\mu$ (up to isomorphism). Composing diagonals and $\mu$, one deduces that $g \mapsto g^n$ goes to itself (up to isomorphism) for any abelian $G$ and integer $n$. Using the classification of finitely generated abelian groups, one deduces that all maps of finite generated abelian groups go to themselves (up to isomorphism).

I believe I was able to argue that the mystery functor must commute with the abelianization function up to natural isomorphism, but I don't have notes from this anymore. My plan for getting beyond this was to chase the isomorphism down the derived series, argue is particular everything must be good for maps of free groups as the intersection of the derived subgroups is trivial there, and then somehow use that everything is a quotient of a free group to win. But I got stuck around this point.

This question has a very similar feel to this math.SE question: Is there a function $\mathcal{P}$ from connected topological spaces to groups such that $\mathcal{P}(X) \cong \pi_1(X)$? One can show that such a functor can't restrict to $\pi_1$ on pointed spaces, which is probably what the OP meant to ask, but it seems hard to rule out that there is some functor.

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