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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeAug 11th 2010
   • (edited Aug 16th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexThe proof of HTT Lemma A.3.6.17 does not seem right. I assume that the lemma is true, but the proof is all mangled, like it was somehow mixed up with an old version. Oddities: It asserts that the fibrant replacement functor induces a functor $F:U_f \to U^\circ$ where $U_f$ is the full subcategory of fibrant objects of $U$, and $U^\circ$ is the full subcategory of fibrant-cofibrant objects. Shouldn't the map classifying the natural transformation $U_f\otimes [1]_S\to U^\circ$ instead have target $U_f$? W'_0 seems like it should be the class of $w\otimes id$ where w is a weak equivalence instead of the class of $e\otimes id$ where e is an equivalence. Questions: Why is it enough to show that the upper and lower inclusion maps are weak equivalences near the end of the proof? Also, how does Corollary A.3.4.11 help in proving it at all? [Link](http://arxiv.org/pdf/math/0608040v4#page=711)

   The proof of HTT Lemma A.3.6.17 does not seem right. I assume that the lemma is true, but the proof is all mangled, like it was somehow mixed up with an old version.

   Oddities:

   It asserts that the fibrant replacement functor induces a functor $F:U_f \to U^\circ$ where $U_f$ is the full subcategory of fibrant objects of $U$, and $U^\circ$ is the full subcategory of fibrant-cofibrant objects. Shouldn’t the map classifying the natural transformation $U_f\otimes [1]_S\to U^\circ$ instead have target $U_f$?

   W’_0 seems like it should be the class of $w\otimes id$ where w is a weak equivalence instead of the class of $e\otimes id$ where e is an equivalence.

   Questions: Why is it enough to show that the upper and lower inclusion maps are weak equivalences near the end of the proof? Also, how does Corollary A.3.4.11 help in proving it at all?

   Link

2. • CommentRowNumber2.
   • CommentAuthorRodMcGuire
   • CommentTimeAug 11th 2010
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   Author: RodMcGuire
   Format: MarkdownItexA link please to what you are talking about, or is your message itself a palimpsest? >is all mangled, like it was somehow mixed up with an old version I think a good term for this pheonomana in math writing is <a href="http://en.wikipedia.org/wiki/Palimpsest#Extended_usages">palimpsest</a> where some old version has not been completely removed and rewritten.

   A link please to what you are talking about, or is your message itself a palimpsest?

   is all mangled, like it was somehow mixed up with an old version

   I think a good term for this pheonomana in math writing is palimpsest where some old version has not been completely removed and rewritten.

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeAug 11th 2010
   • (edited Aug 16th 2010)
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   Author: Harry Gindi
   Format: MarkdownItex-

   -

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeAug 16th 2010
   • (edited Aug 16th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexBump? I did it because Urs and Todd are back from vacation, so I hope they'll have a second to look at it.

   Bump? I did it because Urs and Todd are back from vacation, so I hope they’ll have a second to look at it.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeAug 16th 2010
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   Author: Urs
   Format: MarkdownItexFirst of all, I think you are correctly identifying the typos here. I think instead of "fibrant replacement" we should read "cofibrant replacement". And the "equivalence" that you refer to must be a "weak equivalence", yes. In fact I think there is even more problem with the notation here. For instance $id$ and $F$ and hence $\alpha$ don't really go between $\mathcal{U}_f$ and $\mathcal{U}^\circ$. But as always, all these are pretty trivial things to fix. If one fixes it in the evident way then the main statement remains that we get the map $h$. Let me see about the remaining question about the end of the proof...

   First of all, I think you are correctly identifying the typos here.

   I think instead of “fibrant replacement” we should read “cofibrant replacement”. And the “equivalence” that you refer to must be a “weak equivalence”, yes. In fact I think there is even more problem with the notation here. For instance $id$ and $F$ and hence $\alpha$ don’t really go between $\mathcal{U}_f$ and $\mathcal{U}^\circ$.

   But as always, all these are pretty trivial things to fix. If one fixes it in the evident way then the main statement remains that we get the map $h$.

   Let me see about the remaining question about the end of the proof…

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeAug 16th 2010
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   Author: Urs
   Format: MarkdownItex> Questions: Why is it enough to show that the upper and lower inclusion maps are weak equivalences near the end of the proof? Also, how does Corollary A.3.4.11 help in proving it at all? The map $h$ which we obtain from the fact that we have a functorial cofibrant replacement goes actually $$ h : (\mathcal{U}^\circ \otimes [1]_S)[W'^{-1}] \to \mathcal{U}^\circ[W_0^{-1}] $$ (which incidentally makes the use of "equivalence" instead of "weak equivalence" in the sentence above correct after all), From Lemma A.3.5.14 we can extend this to a span $$ \mathcal{U}^\circ[W_0^{-1}] \otimes [1]_S \stackrel{\simeq}{\leftarrow}(\mathcal{U}^\circ \otimes [1]_S)[W'^{-1}] \stackrel{h}{\to} \mathcal{U}^\circ[W_0^{-1}] \,. $$ To get a homotopy, we need to make this span into a genuine morphism, $$ \mathcal{U}^\circ[W_0^{-1}] \otimes [1]_S \to \mathcal{U}^\circ[W_0^{-1}] \,. $$ i.e. invert the morphism on the left. This is supposedly accomplished by the remaining remarks. Let me see...

   Questions: Why is it enough to show that the upper and lower inclusion maps are weak equivalences near the end of the proof? Also, how does Corollary A.3.4.11 help in proving it at all?

   The map $h$ which we obtain from the fact that we have a functorial cofibrant replacement goes actually

   $$h : (\mathcal{U}^\circ \otimes [1]_S)[W'^{-1}] \to \mathcal{U}^\circ[W_0^{-1}]$$

   (which incidentally makes the use of “equivalence” instead of “weak equivalence” in the sentence above correct after all), From Lemma A.3.5.14 we can extend this to a span

   $$\mathcal{U}^\circ[W_0^{-1}] \otimes [1]_S \stackrel{\simeq}{\leftarrow}(\mathcal{U}^\circ \otimes [1]_S)[W'^{-1}] \stackrel{h}{\to} \mathcal{U}^\circ[W_0^{-1}] \,.$$

   To get a homotopy, we need to make this span into a genuine morphism,

   $$\mathcal{U}^\circ[W_0^{-1}] \otimes [1]_S \to \mathcal{U}^\circ[W_0^{-1}] \,.$$

   i.e. invert the morphism on the left. This is supposedly accomplished by the remaining remarks. Let me see…

7. • CommentRowNumber7.
   • CommentAuthorHarry Gindi
   • CommentTimeAug 16th 2010
   • (edited Aug 16th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexI'm pretty sure that $X\otimes [1]_\mathbf{S}$ is not a cylinder object for an $S$-category $X$ in general because the tensor product does not preserve cofibrations, in this case, $X\coprod X\cong X\otimes [1]_{\emptyset_\mathbf{S}} \to X\otimes [1]_\mathbf{S}$ is not necessarily a cofibration. Also, the thing about equivalences is important for the proof of one direction: When we're looking at the map $U_f\otimes [1]_S \to U_f$, those elements of W' have to be weak equivalences or we don't even have a natural map $U_f[W^{-1}]\to (U_f\otimes [1]_S)[W'^{-1}]$ (on the top or bottom inclusion). Also, lastly, A.3.5.14 says that we have a weak equivalence $C[W^{-1}]\otimes D \to (C\otimes D)[W'^{-1}]$ exactly when $W'^{-1}=\{x\otimes id_D: x\in W\}$. However, in the case of A.3.6.17, there are more morphisms in $W'$ than just those.

   I’m pretty sure that $X\otimes [1]_\mathbf{S}$ is not a cylinder object for an $S$-category $X$ in general because the tensor product does not preserve cofibrations, in this case, $X\coprod X\cong X\otimes [1]_{\emptyset_\mathbf{S}} \to X\otimes [1]_\mathbf{S}$ is not necessarily a cofibration.

   Also, the thing about equivalences is important for the proof of one direction: When we’re looking at the map $U_f\otimes [1]_S \to U_f$, those elements of W’ have to be weak equivalences or we don’t even have a natural map $U_f[W^{-1}]\to (U_f\otimes [1]_S)[W'^{-1}]$ (on the top or bottom inclusion).

   Also, lastly, A.3.5.14 says that we have a weak equivalence $C[W^{-1}]\otimes D \to (C\otimes D)[W'^{-1}]$ exactly when $W'^{-1}=\{x\otimes id_D: x\in W\}$. However, in the case of A.3.6.17, there are more morphisms in $W'$ than just those.

8. • CommentRowNumber8.
   • CommentAuthorHarry Gindi
   • CommentTimeAug 17th 2010
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   Author: Harry Gindi
   Format: MarkdownItexHey Urs, I've been stuck for about a week on this (and this is the last barrier to me finishing chapter 2 of HTT). Not to try to guilt you into answering more quickly, but I would sincerely appreciate any sort of light you could shed on the matter.

   Hey Urs, I’ve been stuck for about a week on this (and this is the last barrier to me finishing chapter 2 of HTT). Not to try to guilt you into answering more quickly, but I would sincerely appreciate any sort of light you could shed on the matter.