Timeline for Abelian category with enough injectives but not functorially
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2020 at 17:07 | vote | accept | Andrea Ferretti | ||
| Apr 8, 2020 at 22:26 | answer | added | R. van Dobben de Bruyn | timeline score: 19 | |
| Apr 8, 2020 at 20:03 | comment | added | Jeremy Rickard | @R.vanDobbendeBruyn Looks good. Thanks. | |
| Apr 8, 2020 at 18:58 | comment | added | R. van Dobben de Bruyn | @JeremyRickard: I think the following is a functorial injective resolution on countable abelian groups. Let $I(A)$ be the quotient of $\mathbf Q^{(A)}$ by the subgroup (not $\mathbf Q$-subspace) generated by $e_{a+b} - e_a - e_b$ for $a,b \in A$. The natural presentation of $A$ as $\mathbf Z^{(A)}$ modulo the same relations gives an injection $A \hookrightarrow I(A)$. Clearly $I(A)$ is divisible and countable, because the same holds for $\mathbf Q^{(A)}$. For example, $I(0) = (\mathbf Q/\mathbf Z)e_0$, and $I(\mathbf Z/2\mathbf Z) = (\mathbf Q/\mathbf Z)e_0 \oplus (\mathbf Q/2\mathbf Z)e_1$. | |
| Apr 7, 2020 at 10:43 | comment | added | Harry Gindi | Welcome back Andrea =] | |
| Apr 7, 2020 at 10:32 | history | edited | Andrea Ferretti | CC BY-SA 4.0 |
deleted 978 characters in body
|
| Apr 7, 2020 at 10:30 | comment | added | Andrea Ferretti | Agreed, I am going to remove the motivation comment | |
| Apr 7, 2020 at 10:28 | comment | added | D.-C. Cisinski | The axiom of choice also holds for classes in the framework of Bernays and Gödel, which is known to be equivalent to ZFC. I think you should delete the last comment from your question because it is misleading, inaccurate, and as such, does distract from the main question. | |
| Apr 7, 2020 at 10:07 | comment | added | Andrea Ferretti | In any case, I wouldn't want this point to distract from the main question, which is about the existence of functorial resolutions - something which is of independent interest | |
| Apr 7, 2020 at 9:56 | comment | added | Andrea Ferretti | In order to obtain from this an actual functor, you need to choose a resolution for each object. Something that you can do with choice on a small category. On a large category, it seems to me to depend on the foundations that you choose, in particular whether you have choice for classes | |
| Apr 7, 2020 at 9:53 | comment | added | Andrea Ferretti | I ensure you that many books do gloss on this point: I checked them before posting. I did not want to say names. I just checked Cartan-Eilbenberg, and it is indeed more precise. I cite straight from Cartan-Eilenberg "Thus up to natural isomorphism HT(X, Y) is independent of the resolutions X and Y and may be written as (RT)(A, C). These modules together with the maps (RT)(A, C) yield a new (additive) functor RT..." Even here, RT(A, C) is determined up to a (uniquely defined) isomorphism. | |
| Apr 7, 2020 at 9:51 | comment | added | D.-C. Cisinski | And this works with very classical set theory with the axiom of choice (no need at all of universes which were not even available at the birth of homological algebra). | |
| Apr 7, 2020 at 9:42 | comment | added | D.-C. Cisinski | The last comment of this post simply not accurate at all: the lack of functoriality of injective resolution does not prevent derived functors to be functorial at all. Saying that other books simply gloss ove this point is simply not true. For instance, the original book of Cartan and Eilenberg does fully explain functoriality of derived functors, and it has been discussed over and over, uing more and more sophisticated tools. | |
| Apr 7, 2020 at 8:28 | comment | added | Andrea Ferretti | This is exactly the problem. I don't have an idea of any criterion on how to prove that a functorial injective embedding does not exist. | |
| Apr 7, 2020 at 8:13 | comment | added | Jeremy Rickard | As a possible candidate for an abelian category with enough injectives, but without functorial injective emveddings, how about the category of countable abelian groups? This has enough injectives (the injective envelope of a countable group is countable), but the functor constructed in the Stacks project gives uncountable injectives. But I have no idea how to prove that there is no other suitable functor. | |
| Apr 7, 2020 at 8:01 | comment | added | Jeremy Rickard | If $I$ were additive then $I(\mathbb{Z}/2\mathbb{Z})$ would have to be $2$-torsion, since $2.\text{id}_{I(\mathbb{Z}/2\mathbb{Z})}=I(2.\text{id}_{\mathbb{Z}/2\mathbb{Z}})=0$. But sure, if $I$ is not additive then it doesn't need to be. | |
| Apr 7, 2020 at 7:39 | history | edited | Andrea Ferretti | CC BY-SA 4.0 |
added 160 characters in body
|
| Apr 7, 2020 at 7:36 | comment | added | Andrea Ferretti | @JeremyRickard Of course you are right, I was silly to require the functor to be additive. I am going to edit that out. That said, I am still not sure why $I(\mathbb{Z}/2\mathbb{Z})$ should be annihilated by $2$. Of course, for all Abelian groups $G$, the image of $\operatorname{Hom}(G, \mathbb{Z}/2\mathbb{Z})$ inside $\operatorname{Hom}(I(G), I(\mathbb{Z}/2\mathbb{Z}))$ is $2$-torsion. But that image could be strictly contained in the Hom group | |
| Apr 6, 2020 at 13:51 | comment | added | Jeremy Rickard | But the functor constructed there is not additive, which was one of the conditions you listed. | |
| Apr 6, 2020 at 13:37 | comment | added | Andrea Ferretti | Well, it would be $\mathbb{Q}/\mathbb{Z}$, right? I don't understand why it would have to be annihilated by $2$. In fact Theorem 19.2.8 in the stacks project constructs a functorial injective embedding for $\operatorname{Mod}_R$. | |
| Apr 6, 2020 at 12:50 | comment | added | Jeremy Rickard | Isn’t the category of abelian groups a counterexample? What could $I(\mathbb{Z}/2\mathbb{Z})$ be? There are no nonzero injective abelian groups annihilated by multiplication by $2$. | |
| Apr 5, 2020 at 17:02 | comment | added | Andrea Ferretti | Plus, I am now actually curious about a possible counterexample | |
| Apr 5, 2020 at 17:02 | comment | added | Andrea Ferretti | I agree, but this is a tad unsatisfying and I think it only works for small categories (here, things can depend whether you take classes in NBG or on a Grothendieck universe) | |
| Apr 5, 2020 at 14:56 | comment | added | Donu Arapura | I don't have an answer to your question, but I agree it probably won't be anything obvious. I do have a comment about the motivation. Even without functorial injective embeddings, one can choose (assuming a strong form of the axiom of choice) an injective resolution for each object of $\mathcal{A}$. The standard arguments, guarantee the existence of a morphism between chosen injective resolutions lifting a given one, and this is unique up to homotopy. This should give functoriality of derived functors. Aside from a philosophical objection to the first step, I'm not sure I see a problem. | |
| Apr 5, 2020 at 11:20 | history | edited | Andrea Ferretti | CC BY-SA 4.0 |
edited body
|
| Apr 5, 2020 at 10:34 | history | asked | Andrea Ferretti | CC BY-SA 4.0 |