Timeline for Homotopy factorization of morphisms of chain complexes
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2014 at 18:43 | comment | added | fosco | Meanwhile: Joyal gave a powerful alternative characterization which give us orthogonality for free if we show that the two classes are closed under isomorphisms and composition and we have a functorial factorization. Maybe the point is that the right procedure to extract the FS from the functorial factorization is not what I said before? | |
| Jun 4, 2014 at 18:27 | comment | added | fosco | MathOverflow wants us to avoid long discussions in comments. Maybe we can continue this discussion in chat? | |
| Jun 4, 2014 at 18:14 | comment | added | Fernando Muro | @tetrapharmakon I must acknowledge that I'm not acquainted enough with the notion of factorisation system. Wikipedia says that an equivalent definition is: the functorial factorization (which is what I give above) + an orthogonality condition. I haven't checked the latter, but it looks plausible. | |
| Jun 4, 2014 at 17:46 | comment | added | fosco | Maybe the point is that it's not really clear to me to which extent a FS is determined by its rule to factor morphisms (the factorization functor sending $f$ to the composable pair $f_E, f_M$ s.t. $f_Mf_E=f$). I can state a precise question if needed. | |
| Jun 4, 2014 at 17:27 | comment | added | Fernando Muro | @tetrapharmakon I just meant that $\mathcal E$ and $\mathcal M$ are not the appropriate classes of morphisms to consider. More conditions are necessary ($A\rightarrow B$ must induce isomorphisms in negative/positive degrees). | |
| Jun 4, 2014 at 16:11 | comment | added | fosco | Oh. I see now that Domenico phrased the original problem in a form weaker than that we need. In fact we were looking for factorization systems on $Ch(R)$, not only for factorization functors (thought as sections of the composition functor $Ch(R)\times Ch(R)\to Ch(R)$)... Can your proof (or Eric's) be adapted to this end? | |
| Jun 4, 2014 at 16:02 | comment | added | fosco | Ha! So this doesn't solve the original problem :( (your answer was extremely useful, though) | |
| Jun 4, 2014 at 15:49 | comment | added | Fernando Muro | @tetrapharmakon I don't think so, too few conditions. | |
| Jun 4, 2014 at 13:24 | comment | added | fosco | I have a similar problem here (see my comment to Eric's answer): now I can define $\mathcal E = \left\{ \begin{smallmatrix} A \\ \downarrow \\ B \end{smallmatrix} \mid \begin{array}{ccc} Z_0 A & \overset{Z_0 f}\to & Z_0 B \\ \downarrow && \downarrow \\ A_0 & \underset{f_0}\to & B_0 \end{array} \text{ is a pushout} \right\}$ and $\mathcal M = \left\{ \begin{smallmatrix} A \\ \downarrow \\ B \end{smallmatrix} \mid \begin{array}{ccc} Z_0 A & \overset{Z_0 f}\to & Z_0 B \\ \downarrow && \downarrow \\ A_0 & \underset{\cong}\to & A\amalg_{ Z_0A}Z_0B \end{array} \right\}$. Is it a FS? | |
| Jun 2, 2014 at 23:04 | history | answered | Fernando Muro | CC BY-SA 3.0 |