Timeline for Filling out Grothendieck's proof that effaceable $\delta^*$-functors are universal
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 16, 2021 at 6:36 | comment | added | Aryaman Maithani | @D.Wynter: That is shown in this answer. | |
| May 20, 2019 at 11:58 | comment | added | user219967 | @CaveJohnson Thanks a lot, I understand your solution. | |
| Oct 14, 2018 at 14:12 | comment | added | Cave Johnson | Commutativity with connecting maps can be shown in a simpler way: for $0\to A\stackrel f\to B\to C\to0$ an exact sequence and $u_B:B\hookrightarrow M_B$ an effacement, $u_B\circ f:A\hookrightarrow M_B$ is also an effacement, making $$\begin{array}[ccccc]{} 0&\to&A&\stackrel f\to&B\\&&\|&&\downarrow u_B\\0&\to&A&\stackrel{u_B\circ f}\to&M_B\end{array}$$ commutative. This diagram extends to a morphism between the exact sequences. Passing to long exact sequences, we can decompose connecting maps of the arbitrary exact sequence by maps of effacement exact sequence, which we've already worked out. | |
| Jan 26, 2017 at 1:35 | comment | added | Ben Webster♦ | If $M_A$ and $\tilde{M}_A$ are both possibilities, you show that the natural map $A\to M_A\oplus \tilde{M}_A/\{a,-a)\mid a\in A\}$ also effaces, and then use the maps induced by the inclusions of $M_A$ and $\tilde{M}_A$ into this bigger object. | |
| Jan 25, 2017 at 3:24 | comment | added | Dominic Wynter | But how do we know that the map $f_A^n$ is independent of the choice of $M_A$? | |
| Jan 25, 2017 at 3:21 | history | answered | Ben Webster♦ | CC BY-SA 3.0 |