Timeline for Trees in chain complexes
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 13, 2020 at 16:08 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
added 343 characters in body
|
| Oct 13, 2020 at 5:51 | comment | added | D.-C. Cisinski | Prop. 2.15 and 2.20 from this paper of mine numdam.org/item/BSMF_2010__138_3_317_0 tell us that the functor $Ho(Fun(T,M))\to Fun(T,Ho(M))$ full, essentially surjective and conservative for any model category $M$ and any tree $T$. Hence it induces a bijection on the collections of isomorphism classes. | |
| Oct 13, 2020 at 3:42 | comment | added | Dmitri Pavlov | @TimCampion: Thanks, I threw away the claim about π_1 (which stopped making sense the moment I reread it), just left the claim for π_0 (which matches the original question). | |
| Oct 13, 2020 at 3:38 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
deleted 1126 characters in body
|
| Oct 13, 2020 at 2:52 | comment | added | Tim Campion | More explicitly, cofibrantly replace $F$ by $F' = (k[n] \to D[n+1])$. Then $Hom(F',G)$ is 1-dimensional and $Hom(\Sigma F', G) = 0$, so the only nullhomotopy of maps $F' \to G$ is zero. Thus $RHom(F,G) = Hom(F',G)$ is 1-dimensional whereas $Hom(F,G) = 0$, so again the former is not a quotient of the latter. | |
| Oct 13, 2020 at 2:46 | comment | added | Tim Campion | Thanks for the details. I'm still skeptical. Let $T$ be the arrow category and let $F = (k[n] \to 0)$ and $G = (0 \to k[n+1])$. Then $RHom(F , G)$ is the space of commutative squares $k[n] \rightrightarrows 0,0 \rightrightarrows k[n+1]$, i.e. the space of maps $k[n] \to \Omega k[n+1] = k[n]$, which is 1-dimensional. But $Hom(F,G) = 0$. So $RHom(F,G)$ is not a quotient of $Hom(F,G)$. | |
| Oct 13, 2020 at 0:36 | comment | added | Dmitri Pavlov | @TimCampion: I don't even know why I mentioned Dwyer–Kan equivalences in the first place, since I never use them. I added details for the other issue. | |
| Oct 13, 2020 at 0:35 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
added 1031 characters in body
|
| Oct 12, 2020 at 22:04 | comment | added | Tim Campion | Oh I think I understand now how you're trying to avoid needing naturality. The issue now becomes that $\iota H(F)$ is not cofibrant in general, so it's not the case in general that $Hom_{Ho(Fun(T,Ch_k))}(\iota H(F), \iota H(G))$ is a quotient of $Hom_{Fun(T,Ch_k)}(\iota H(F),\iota H(G))$ by a congruence. | |
| Oct 12, 2020 at 20:08 | vote | accept | Surojit Ghosh | ||
| Oct 12, 2020 at 20:05 | comment | added | Tim Campion | Also, your second paragraph, you claim that $\iota, H$ are DK-equivalences. I don't think this is true -- $Gr_k$ is a 1-category, but the DK-localization of $Ch_k$ has higher homotopy in its hom-spaces (given by degree-shifting maps). | |
| Oct 12, 2020 at 19:38 | comment | added | Tim Campion | Thanks! I think we're doing similar things. I don't see why these isomorphisms are natural though, so I'm not sure I buy the claim that we have an equivalence of categories $Ho(Fun(T,Ch_k)) \simeq Fun(T,Ho(Ch_k))$. | |
| Oct 12, 2020 at 19:26 | comment | added | Dmitri Pavlov | @TimCampion: I added a new paragraph explaining this. | |
| Oct 12, 2020 at 19:26 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
added 421 characters in body
|
| Oct 12, 2020 at 19:12 | comment | added | Tim Campion | I assume that $H: Ch_k \to Ho(Ch_k) = Gr_k$ is the homology functor and that $\iota : Ho(Ch_k) \to Ch_k$ is its "zero-differential" section. In the second-to-last paragraph, I don't understand where you get a map $Hom(F,G)/\sim \to Hom(\iota H(F), \iota H(G))/\sim$ from -- the quasi-isomorphisms between $F(v)$ and $\iota HF(v)$ are not natural for $v \in T$, so it's not immediately obvious that these levelwise maps can be made into natural maps between $F$ and $G$. | |
| Oct 12, 2020 at 19:05 | vote | accept | Surojit Ghosh | ||
| Oct 12, 2020 at 20:07 | |||||
| Oct 12, 2020 at 18:54 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |