Timeline for Quillen equivalence for under-categories
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2018 at 18:32 | history | edited | Valery Isaev | CC BY-SA 3.0 |
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| Apr 21, 2017 at 4:55 | history | edited | Valery Isaev | CC BY-SA 3.0 |
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| Apr 21, 2017 at 3:53 | comment | added | Victor | Let us continue this discussion in chat. | |
| Apr 21, 2017 at 3:33 | history | edited | Valery Isaev | CC BY-SA 3.0 |
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| Apr 21, 2017 at 3:28 | comment | added | Valery Isaev | Ah, yes, you are right. But it still seems that it is not enough to assume that $c \to R(d)$ is a weak equivalence. | |
| Apr 21, 2017 at 3:17 | comment | added | Victor | $c_1$ is not cofibrant and thus the composition of the first two maps does not have to be a weak equivalence. | |
| Apr 21, 2017 at 2:49 | comment | added | Valery Isaev | No, I didn't. If $\hat{L} \dashv \hat{R}$ is a Quillen equivalence, then for every cofibration $c \to c_1$, the map $c_1 \to RL(c_1) \to RSL(c_1) \to RS(d_1)$ is a weak equivalence (this is a part of a characterization of Quillen equivalences). The composition of the first two maps is a weak equivalence and $R$ reflects weak equivalences between fibrant objects (this follows from the fact that $L \dashv R$ is a Quillen equivalence). $S$ also reflect weak equivalences. Thus $L(c_1) \to d_1$ must be a weak equivalence if $\hat{L} \dashv \hat{R}$ is a Quillen equivalence. | |
| Apr 21, 2017 at 2:44 | comment | added | Victor | But you also assume that $L(c)\to d$ is a weak equivalence, while in my case it is not true: $c\to R(d)$ is! | |
| Apr 21, 2017 at 2:40 | comment | added | Valery Isaev | Well, the proof shows that this is the case for maps $L(c_1) \to d_1$ and not for adjoint ones. | |
| Apr 21, 2017 at 2:34 | comment | added | Victor | I believe that $\hat L\dashv\hat R$ is a Quillen equivalence if and only if for every cofibration $c\to c_1$, the map $c_1\to R(d_1)$ (the adjoint of your map) is a weak equivalence. I reiterate I need this result when $c$ is not necessarily cofibrant, thus it is not equivalent saying that $L(c_1)\to d_1$ is a weak equivalence, which is not true in my case. | |
| Apr 21, 2017 at 2:32 | comment | added | Valery Isaev | Also, if $c$ is cofibrant and $c \to R(d)$ is a weak equivalence, then so is $L(c) \to d$. | |
| Apr 21, 2017 at 2:19 | comment | added | Valery Isaev | The condition that $L(c) \to d$ is a weak equivalence is actually necessary. Indeed, the map $L(c_1) \to d_1$ that I constructed in the proof is a weak equivalence for every cofibration $c \to c_1$ if and only if $\hat{L} \dashv \hat{R}$ is a Quillen equivalence. But if we take $id_c : c \to c$, then this map is simply $L(c) \to d$. | |
| Apr 21, 2017 at 2:11 | comment | added | Victor | In the case I am interested in, $L(c)\to d$ is not a weak equivalence, only $c\to R(d)$ is. | |
| Apr 21, 2017 at 0:30 | history | answered | Valery Isaev | CC BY-SA 3.0 |