Timeline for recognising weak equivalences of simplicial sets
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30, 2015 at 1:31 | vote | accept | COhrt | ||
| Jan 29, 2015 at 21:26 | comment | added | Dmitri Pavlov | The edit also addresses the problem that I pointed out in the above comment. | |
| Jan 29, 2015 at 18:04 | comment | added | Karol Szumiło | You are right, that's my bad. I have missed a step, but I have edited my answer to fix it. | |
| Jan 29, 2015 at 18:04 | history | edited | Karol Szumiło | CC BY-SA 3.0 |
added 349 characters in body
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| Jan 29, 2015 at 17:16 | comment | added | COhrt | Then I don't understand how in your first step it is enough to only show that $Ex^\infty f$ admits a lift in the squares factorizing through $X$ and $Y$ to show that it is a weak equivalences. Wouldn't we have to show that $\Ex^\infty f$ admits a lift in all squares involving $\partial \Delta^n\to \Delta^n$? | |
| Jan 29, 2015 at 16:58 | comment | added | Karol Szumiło | That's not true and I'm not using that, I'm only using that there is a factorization $\partial\Delta[n] \to \mathrm{Ex}^k X \to \mathrm{Ex}^\infty X$ for some $k$. | |
| Jan 29, 2015 at 16:43 | comment | added | COhrt | Thank you, but your answer seems to assume that every map $\partial\Delta^n\to {Ex}^\infty X$ factorizes as $\partial\Delta^n\to X\to {Ex}^\infty X$ and that this factorization is functorial in some sense. Why is this always true? | |
| Jan 29, 2015 at 15:44 | comment | added | Karol Szumiło | A relative lift involves a choice of a homotopy, so my $k$ is already your $l$. | |
| Jan 29, 2015 at 15:37 | comment | added | Dmitri Pavlov | But the homotopy need not factor through the same Ex^k, it might land in some Ex^l where l is much bigger than k. So in the lifting condition one must allow k to increase first. | |
| Jan 29, 2015 at 12:57 | history | answered | Karol Szumiło | CC BY-SA 3.0 |