Timeline for Abelian category with enough injectives but not functorially
Current License: CC BY-SA 4.0
13 events
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| Apr 29, 2020 at 5:03 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Simplified representation-theoretic argument.
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| Apr 9, 2020 at 17:07 | vote | accept | Andrea Ferretti | ||
| Apr 9, 2020 at 16:39 | comment | added | Peter LeFanu Lumsdaine | @JeremyRickard: You’re right, that functoriality case (with the maps from $\mathbb{Z}$) does fail — I’d been sloppy in thinking it through! So the “obstacle” is not a problem; sorry for the noise. (The action on finite groups does work fine, though, that I did check carefully: it’s perhaps clearer if you present $F(A)$ as $\mathbb{Z}^{\oplus A}/(e_0)$ rather than $\mathbb{Z}^{\oplus (A \setminus \{0\})}$. The reason for killing $e_0$, not just using $\mathbb{Z}^{\oplus A}$, was because I thought that would allow the functor to be extended to a fin. gen. value on $\mathbb{Z}$.) | |
| Apr 9, 2020 at 16:25 | comment | added | Andrea Ferretti | You are right, and I agree that $f(a) = 0$ is an issue, but then one could just take the free group over $A$. Your previous example of composing with a map from $\mathbb{Z}$ is where functoriality seems to fail | |
| Apr 9, 2020 at 16:15 | comment | added | Jeremy Rickard | @AndreaFerretti I don't think being a homomorphism is a problem: we're just defining a map between free abelian groups by saying where a basis gets sent. But there is a problem: what happens if $f(a) = 0$? | |
| Apr 9, 2020 at 16:08 | comment | added | Andrea Ferretti | $F$ does not need to be additive, but $F(f)$ had better be a homomorphism | |
| Apr 9, 2020 at 16:06 | comment | added | Jeremy Rickard | @AndreaFerretti I thought we didn't want $F$ to be additive? | |
| Apr 9, 2020 at 15:16 | comment | added | Andrea Ferretti | @PeterLeFanuLumsdaine I also do not understand the action of $F$ on maps between finite groups. If $A$ and $B$ are finite and $f \colon A \to B$, the only obvious candidate for $F(f)$ would send the generator $a \in F(A)$ to the generator $f(a) \in F(B)$, but that is not additive | |
| Apr 9, 2020 at 14:50 | comment | added | Jeremy Rickard | @PeterLeFanuLumsdaine I don't see how you get a functor, but maybe "the first thing I guess" is not what you thought would be!. If you take the map $\mathbb{Z}\to\mathbb{Z}$ given by multiplication by $2n$ and the nonzero map $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$, the composition is zero, independently of $n$. How do you define the functor on these maps? | |
| Apr 9, 2020 at 10:01 | comment | added | Peter LeFanu Lumsdaine | The functor is given as follows: For finite $A$, take $F(A)$ to be free on the set $A \setminus \{0\}$; this is easily seen to be a functorial f.g. projective cover. Take $F(\mathbb{Z})$ to be $\mathbb{Z}$ with the identity cover. The action on maps to/from $\mathbb{Z}$ is “the first thing you guess” in each case; checking functoriality has to work by cases depending whether the source/target of maps are $\mathbb{Z}$ or finite, but all cases work out, essentially since there are no nonzero maps from finite groups to $\mathbb{Z}$. Am I mistaken, or does your answer get around this somehow? | |
| Apr 9, 2020 at 9:53 | comment | added | Peter LeFanu Lumsdaine | When considering this question the other day, I had convinced myself there was an obstacle to approaches like this one, and I don’t see how this answer gets around the obstacle (though I don’t see any flaw in your argument either). The obstacle is this: It seems that after the use of Lemma 2, the proof only uses the value of $F$ on cyclic groups. But if I’m not mistaken, there is a functor defined on $\mathbb{Z}$ and all finite groups giving a faithful f.g. projective cover (details in next comment). So any answer must use values of $F$ on non-cyclic infinite groups. | |
| Apr 9, 2020 at 1:20 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Explained idea of proof.
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| Apr 8, 2020 at 22:26 | history | answered | R. van Dobben de Bruyn | CC BY-SA 4.0 |