Timeline for Coequalizers in stable (infinity,1)-categories
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2022 at 5:20 | comment | added | Ken | To show that $C$ is the coequalizer of $f$ and $g$, you could also use HTT, Proposition 4.4.2.2 (to $\Delta ^0 \amalg _{\partial\Delta^{\{0,1\}}}\Lambda^2_2$). | |
| Jul 30, 2013 at 20:43 | vote | accept | Mike Shulman | ||
| Jul 29, 2013 at 16:19 | comment | added | Omar Antolín-Camarena | Sorry @MikeShulman, I hadn't seen your last remark. That's right: the inclusion of the 4-simplices glued together into $\Delta^1 \times \Delta^2$ is just $\Delta^1$ cross the inclusion of $\Lambda^2_1$ into $\Delta^2$. | |
| Jul 23, 2013 at 19:14 | comment | added | Mike Shulman | Of course, $\Delta^1\times \Delta^2$ is more than just four 2-simplices, but I guess four 2-simplices includes into it by an anodyne map or something? | |
| Jul 23, 2013 at 18:14 | comment | added | Omar Antolín-Camarena | To construct the diagrams (which indeed are supposed to be maps $\Delta^1 \times \Delta^2 \to \mathcal{C}$, I'd give four 2-simplices in $\mathcal{C}$ that share sides appropriately. Proving those 2-simplices exist is a hassle involving the definition of $-id$. The worst is probably the one showing that $\pmatrix{0&1\\1&-1} \pmatrix{1\\1} = \pmatrix{1\\0}$. Is that what you had in mind? | |
| Jul 23, 2013 at 17:55 | comment | added | Omar Antolín-Camarena | To show that $C$ is the coequalizer, I'd say that you just take mapping spaces into an arbitrary object $Z$ and that reduces it to showing that for spaces a homotopy equalizer of two maps $p,q:A\to B$ can be computed as the homotopy pullback of $B \xrightarrow{diag} B \times B \xleftarrow{(p,q)} A$. | |
| Jul 23, 2013 at 17:50 | comment | added | Omar Antolín-Camarena | You're certainly right about the $h(f-g)$ proof in an Abelian category, @MikeShulman, I shouldn't have said that first sentence. | |
| Jul 23, 2013 at 17:09 | comment | added | Mike Shulman | As for this proof, how do you show that $C$ being a pushout of that square is equivalent to being the coequalizer? Also, I think I know the answer to this, but how do you actually obtain all of these diagrams (which presumably are supposed to be maps $\Delta^1 \times \Delta^2\to \mathcal{C}$ or something)? | |
| Jul 23, 2013 at 17:03 | comment | added | Mike Shulman | I don't think you need to do anything this complicated in an ordinary abelian category; can't you just say that since composition distributes over addition, a map $h:Y\to Z$ satisfies $h f = h g$ iff it satisfies $h (f - g) = 0$, so the two have the same universal property? | |
| Jul 22, 2013 at 19:07 | history | edited | Omar Antolín-Camarena | CC BY-SA 3.0 |
added Akhil's proof of the missing pushout
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| Jul 22, 2013 at 18:45 | comment | added | Akhil Mathew | Hi Omar. I think you can use the "shearing" isomorphism of $X \oplus X$ to turn the sum map $X \oplus X \to X$ into the map $(1, 0): X \oplus X \to X$, and now the map $d$ turns into $(0, 1): X \to X \oplus X$, so the left-hand exact triangle becomes a sum of two (simpler) exact triangles. | |
| Jul 22, 2013 at 18:21 | history | edited | Omar Antolín-Camarena | CC BY-SA 3.0 |
added confesion about left square
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| Jul 22, 2013 at 17:54 | history | answered | Omar Antolín-Camarena | CC BY-SA 3.0 |