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Show $\mathbf{Ab}$ (the category of Abelian Groups) is not equivalent to its opposite category.

I think that I have to show a property such that the dual property is not satisfied. But I am having trouble finding one.

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  • $\begingroup$ Hmm... for any abelian group $G$, there is an epimorphism of a coproduct of some number of copies of $\mathbb{Z}$ to $G$. So the image $A$ of $\mathbb{Z}$ under the equivalence would have to be fairly strange, in order for there to be a monomorphism of any abelian group $G$ to a product of some number of copies of $A$. $\endgroup$ – Daniel Schepler Feb 23 '18 at 1:48
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    $\begingroup$ $\operatorname{Hom}(\mathbb{Z}, -)$ is faithful since it's isomorphic to the forgetful functor. So if you could show there is no abelian group $A$ such that $\operatorname{Hom}(-, A)$ is faithful, that would do the trick... Not sure if that's true, though. $\endgroup$ – Daniel Schepler Feb 23 '18 at 2:02
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    $\begingroup$ @DanielSchepler: Wouldn't $A = \mathbb{Q}/\mathbb{Z}$ give a faithful functor? $\endgroup$ – user14972 Feb 23 '18 at 3:23
  • $\begingroup$ @Hurkyl I agree, $\mathbb{Q}/\mathbb{Z}$ is a cogenerator. $\endgroup$ – Ishan Levy Feb 23 '18 at 6:45
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The identity functor on Ab is representable by $\mathbb{Z}$ but is not corepresentable. Indeed, if $X$ were a corepresenting object, $X \cong $ Hom$(\mathbb{Z},X) \cong \mathbb{Z}$, but certainly $\mathbb{Z}$ isn't the corepresenting object. Thus Ab isn't the same as its dual.

Edit: Since I was asked for a simple proof not using representability, here is another:

A contravariant equivalence $F\colon $ Ab $\to $ Ab sends simple objects to simple objects since Ab is abelian. Thus the simple objects ($\mathbb{Z}/p\mathbb{Z}$) are permuted in such an equivalence. Since End$(\mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z}$, these objects are actually fixed.

Now, the exact sequence $0 \to \mathbb{Z} \xrightarrow{p} \mathbb{Z} \to \mathbb{Z}/p\mathbb{Z} \to 0$ gets sent to $0 \to \mathbb{Z}/p\mathbb{Z} \to F(\mathbb{Z}) \xrightarrow{p} F(\mathbb{Z}) \to 0$, so the subgroup of $F(\mathbb{Z})$ killed by $p$ is $\mathbb{Z}/p\mathbb{Z}$, and $F(\mathbb{Z})$ is moreover divisible since multiplication by $p$ is surjective for each $p$. It follows that $F(\mathbb{Z})$ is $\mathbb{Q}/\mathbb{Z} \oplus \mathbb{Q}^I$, but then End$(\mathbb{Q}/\mathbb{Z}\oplus \mathbb{Q}^I) \neq$ End$(\mathbb{Z})^{op} = \mathbb{Z}$, a contradiction.

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  • $\begingroup$ Would you have to put some co-abelian group structure on $\mathbb{Z}$ in order for this to make sense? $\endgroup$ – Daniel Schepler Feb 23 '18 at 1:50
  • $\begingroup$ No, because the category is self-enriched anyway. $\endgroup$ – Ishan Levy Feb 23 '18 at 1:52
  • $\begingroup$ Do you think there is a simpler answer not using representability? $\endgroup$ – 300days Feb 23 '18 at 1:54
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    $\begingroup$ Yeah, was just about to say that :) $\endgroup$ – Ishan Levy Feb 23 '18 at 6:46
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    $\begingroup$ @DanielSchepler In other words, every abelian group has a canonical structure of a co-abelian group object in abelian groups. $\endgroup$ – Kevin Arlin Feb 23 '18 at 21:14
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Here's yet another way to show it. In $Ab$, any coproduct of injective objects is injective. Concretely, the injective objects of $Ab$ are the divisible abelian groups, and any direct sum of divisible abelian groups is still divisible.

The dual property is that any product of projective objects is projective. This is false in $Ab$; for instance, a product $\mathbb{Z}^\mathbb{N}$ of infinitely many copies of $\mathbb{Z}$ is not projective (see Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?).


And another, related to Hurkyl's answer. Every finitely generated abelian group $A$ is compact, meaning that the Hom-functor $\operatorname{Hom}(A,-)$ preserves filtered colimits, However, no nontrivial abelian group satisfies the dual version of compactness (that $\operatorname{Hom}(-,A)$ turns filtered limits into filtered colimits); see Can a nonzero module be cocompact?. So, for instance, $Ab^{op}$ has the property "every compact object is terminal" but $Ab$ does not.

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With some knowledge of homological algebra....

One of the important properties of Ab is that direct colimits commute with finite limits. Inverse limits, however, do not generally commute with finite colimits.

Since these properties are dual, the opposite is true of $\mathbf{Ab}^{\mathrm{op}}$; in this category inverse limits commute with finite colimits, but direct colimits do not generally commute with finite limits.

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  • $\begingroup$ I wouldn't call this homological algebra. All you wrote is in MacLane's book, for example. :-) $\endgroup$ – Mariano Suárez-Álvarez Feb 23 '18 at 6:47
  • $\begingroup$ Direct colimits only commute with finite limits! For instance, an infinite product of copies of an infinite torsion group which has elements of unbounded order (such as a Prüfer group) will not be torsion, while the union of infinite products of its finite subgroups will. $\endgroup$ – Kevin Arlin Feb 23 '18 at 21:04
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    $\begingroup$ There's a very cool related theorem of Gabriel and Ulmer: no non-small locally presentable category (which is the direct colimits commute with finite limits thing, plus a few more properties) has locally presentable opposite, so in particular no such category is self-dual. $\endgroup$ – Kevin Arlin Feb 23 '18 at 21:17
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If $F$ is such an anti-equivalence, $A=F(\mathbb Z)$ is an injective abelian group, that is, a divisible abelian group, with endomorphism ring isomorphic to $\mathbb Z$. In particular, $A$ is indecomposable, so it is isomorphic to one of the possible direct summands of a divisible abelian group described in the Structure Theorem of such groups. It turns out that none of them have that endomorphism ring.

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  • $\begingroup$ This proof is very similar to my second one, just that more of the structure theory is used instead of computing $F(\mathbb{Z}/p\mathbb{Z})$. $\endgroup$ – Ishan Levy Feb 23 '18 at 6:57
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    $\begingroup$ One could alternatively do the following: as you have shown, the $F(Z/pZ)$ are permuted, and as $Z$ surjects onto each $Z/pZ$, $F(Z/pZ)$ injects into $F(Z)$. This means that $F(Z)$ has elements of all prime orders and, since it is a direct sum of its primary parts, it is decomposable: this does not play well with the fact that its endomorphism ring is Z. This does not need that one know the form of injectives. $\endgroup$ – Mariano Suárez-Álvarez Feb 23 '18 at 7:17
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Here is, once again, another way to do this: in $\text{Ab}$, or more generally in any abelian category with infinite coproducts and products, there is a canonical map

$$\bigoplus_{i \in I} A_i \to \prod_{i \in I} A_i$$

from an infinite coproduct to an infinite product, and it is a monomorphism but generally not an epimorphism (if all but finitely many of the $A_i$ are nonzero). However, in $\text{Ab}^{op}$ the same map is just the opposite of this map, so it's an epimorphism but generally not a monomorphism.

This observation has come up before on math.SE; see, for example, this answer.

A thing I repeatedly mention is that by Pontryagin duality $\text{Ab}^{op}$ is equivalent to the category of compact Hausdorff abelian groups, which gives a way to see what's going on with the above map in $\text{Ab}^{op}$: the product of compact Hausdorff abelian groups is the cartesian product but the coproduct is the Bohr compactification of the direct sum (which is not compact) which turns out to be "bigger" than the product.

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