# Is this the free abelian group functor?

Let $\mathbb{Z}(.) : \mathbf{Set} \to \mathbf{Ab}$ be the functor that assigns to any set $S$ the set of maps $\mathbb{Z}(S) := \{ z: S \to \mathbb{Z} \; | \; z(s)=0 \mbox{ for almost all } s \in S \}$ and to any set map $f: S \to T$ the morphism $\mathbb{Z} f :\mathbb{Z}(T) \to \mathbb{Z}(S)$ of abelian groups, defined by $\mathbb{Z} f(z):= z \circ f$ for all $z \in \mathbb{Z}(T)$.

This defines a contravariant functor. Right?

Is this what is called the free abelian group functor? (I wonder because of its contravariance)

• As a general rule, free functors are covariant; they are usually left adjoints to other covariant functors (e.g. the forgetful functor $\text{Ab} \to \text{Set}$ in this case). – Qiaochu Yuan Jun 7 '12 at 0:58

## 1 Answer

What you've defined is not a functor, covariant or contravariant. Let $S$ be an infinite set and $f : S \to 1$ the unique map. Then $\mathbb{Z} f$ does not exist. (You need to require that $f$ is proper, that is, that the preimage of a finite set is finite.)

The free abelian group functor is covariant. You've assigned the right things to objects (more or less) but not to morphisms. The free abelian group functor assigns to a set $S$ the abelian group $\mathbb{Z}[S]$ of formal linear combinations $$\sum_{s \in S} c_s s, c_s \in \mathbb{Z}$$

and assigns to a function $f : S \to T$ the homomorphism $\mathbb{Z}[f]$ sending $\sum c_s s$ to $\sum c_s f(s)$. In this setup, the homomorphism you wanted to assign to a function $g : T \to S$ sends $\sum c_s s$ to $$\sum c_s \sum_{g(t) = s} t$$

and of course this is not well-defined if $\{ t : g(t) = s \}$ is infinite. This desire to "integrate over" inverse images appears in other contexts (e.g. pullbacks in homology, of which this may be regarded as a toy example (the $H_0$ of discrete spaces)) but I am not well-qualified to discuss them.

This issue is precisely the reason why I get annoyed when people call $\mathbb{Z}^S$ the free abelian group on $S$ when $S$ is finite: the assignment $S \to \mathbb{Z}^S$ ought to be contravariant, not covariant.

Edit: The comparison to homology might be valuable as a way of contextualizing this discussion. If $S$ is a set regarded as a discrete space, $\mathbb{Z}[S]$ is the zeroth homology $H_0(S)$ while $\mathbb{Z}^S$ is the zeroth cohomology $H^0(S)$; in particular, the former is covariant while the latter is contravariant. The fact that for $S$ finite we can identify the two can then be thought of as a very special case of Poincaré duality, which hammers home the point that the finiteness of $S$ is essential here.

• There are a few situations where there are several natural functors that are the same on objects but different on arrows, so there's always going to be a little irritation from notation that emphasizes objects. Of course, the fact we have perfectly reasonable notation $\mathbb{Z}S$ makes $\mathbb{Z}^S$ much more questionable. – Hurkyl Jun 7 '12 at 0:53
• Ah ok. So my functor only makes sense on the category of finite sets. – Mark Neuhaus Jun 7 '12 at 1:25
• But wat do you mean by "you have assigned the right thing to objects more or less". Why more or less? --- And by the way: I like the (co)homology example – Mark Neuhaus Jun 7 '12 at 1:27
• @Mark: because I am picky. The group you wrote down is of course isomorphic to $\mathbb{Z}[S]$ and there is even a distinguished such isomorphism but, as I already said, I don't like this description of $\mathbb{Z}[S]$ precisely because it suggests contravariance when you want covariance. – Qiaochu Yuan Jun 7 '12 at 1:37
• Ok. As it turns out I want contravariance (not the free functor), so I should use another symbol instead of $\mathbb{Z}[.]$ and of course restrict to finite sets. – Mark Neuhaus Jun 7 '12 at 2:26