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1. • CommentRowNumber1.
   • CommentAuthorYaron
   • CommentTimeMay 18th 2012
   • (edited May 19th 2012)
   • PermaLink
   Author: Yaron
   Format: MarkdownItexI got stuck on something that should probably be obvious in Section VII.6 in CWM (the bar construction). If $A$ is an $Ab$-category and $\langle L,\varepsilon\colon L\Rightarrow I_A, \nu\colon L\Rightarrow L^2\rangle$ is a comonad in $A$ (that is, a comonoid in the strict monoidal category $A^A$), then because the opposite (augmented) simplicial category $\Delta^{\mathrm{op}}$ has the universal comonoid, there is a unique strict monoidal functor $\Delta^{\mathrm{op}}\to A^A$ with $1\mapsto L$, $(\delta_{0}^{0})^{\mathrm{op}}\mapsto \varepsilon$ and $(\sigma_0^1)^{\mathrm{op}}\mapsto \nu$ (using the indexing convention of CWM for face maps $\delta_{?}^{?}$ and degeneracy maps $\sigma_{?}^{?}$ in $\Delta$). Composing with the ``evaluate at $a$'' functor $E_a\colon A^A\to A$, we get an augmented simplical object $Y_a\colon\Delta^{\mathrm{op}}\to A$ with $Y_a(n)=L^n(a)$. As usual, this defines a chain complex $$ (Y_a(0)=a)\leftarrow (Y_a(1)=La)\leftarrow (Y_a(2)=L^2a)\leftarrow\cdots $$ in $A$ (where the boundary morphisms are the appropriate sums of face maps with alternating sums). Now, at the bottom of p. 181 (second edition), it is written the this complex is a resolution. Why is this true in this general setting? [For the special case of the bar resolution appearing in any homology book (that is, when $A={\Pi}$-$\mathbf{Mod}$ for some group ${\Pi}$, $L=\mathbb{Z}({\Pi})\otimes-$, $a=\mathbb{Z}$, etc.), an explicit contracting homotopy is specified (by elements). The problem is that I don't understand how to deal with the general case.]

   I got stuck on something that should probably be obvious in Section VII.6 in CWM (the bar construction).

   If $A$ is an $Ab$-category and $\langle L,\varepsilon\colon L\Rightarrow I_A, \nu\colon L\Rightarrow L^2\rangle$ is a comonad in $A$ (that is, a comonoid in the strict monoidal category $A^A$), then because the opposite (augmented) simplicial category $\Delta^{\mathrm{op}}$ has the universal comonoid, there is a unique strict monoidal functor $\Delta^{\mathrm{op}}\to A^A$ with $1\mapsto L$, $(\delta_{0}^{0})^{\mathrm{op}}\mapsto \varepsilon$ and $(\sigma_0^1)^{\mathrm{op}}\mapsto \nu$ (using the indexing convention of CWM for face maps $\delta_{?}^{?}$ and degeneracy maps $\sigma_{?}^{?}$ in $\Delta$). Composing with the “evaluate at $a$” functor $E_a\colon A^A\to A$, we get an augmented simplical object $Y_a\colon\Delta^{\mathrm{op}}\to A$ with $Y_a(n)=L^n(a)$.

   As usual, this defines a chain complex

   $$(Y_a(0)=a)\leftarrow (Y_a(1)=La)\leftarrow (Y_a(2)=L^2a)\leftarrow\cdots$$

   in $A$ (where the boundary morphisms are the appropriate sums of face maps with alternating sums).

   Now, at the bottom of p. 181 (second edition), it is written the this complex is a resolution. Why is this true in this general setting?

   [For the special case of the bar resolution appearing in any homology book (that is, when $A={\Pi}$-$\mathbf{Mod}$ for some group ${\Pi}$, $L=\mathbb{Z}({\Pi})\otimes-$, $a=\mathbb{Z}$, etc.), an explicit contracting homotopy is specified (by elements). The problem is that I don’t understand how to deal with the general case.]

2. • CommentRowNumber2.
   • CommentAuthorFinnLawler
   • CommentTimeMay 18th 2012
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   Author: FinnLawler
   Format: MarkdownItex[This nCafé post](http://golem.ph.utexas.edu/category/2007/05/on_the_bar_construction.html) by Todd might help you.

   This nCafé post by Todd might help you.

3. • CommentRowNumber3.
   • CommentAuthorYaron
   • CommentTimeMay 18th 2012
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   Author: Yaron
   Format: MarkdownItexThank you very much, Finn. I will surely look into this (although from a first glance it seems a little above my current knowledge). Is there some simple straightforward explanation (something like: "just take ... as the contracting homotopy")?

   Thank you very much, Finn. I will surely look into this (although from a first glance it seems a little above my current knowledge). Is there some simple straightforward explanation (something like: “just take … as the contracting homotopy”)?

4. • CommentRowNumber4.
   • CommentAuthorFinnLawler
   • CommentTimeMay 19th 2012
   • PermaLink
   Author: FinnLawler
   Format: MarkdownItexI don't really know complexes, but for simplicial objects the idea is that if you have a monad $T$ on a category $C$, then the bar construction turns each $T$-algebra into a simplicial $T$-algebra, using the comonad $F^T U^T$ on $C^T$ arising from $T$. If you apply $U^T$ to this simplicial algebra you get a simplicial object of $C$, and it turns out that the unit of $T$ supplies a contracting homotopy for this. (Compare the example at [[split coequalizer]].) Mac Lane says that the complex $L^* a$ is a resolution, even though the associated simplicial object of $A$ won't be contractible in general. Maybe it's exact for some other reason, but you'll have to ask someone who knows more homological algebra than I do (i.e. pretty much anyone).

   I don’t really know complexes, but for simplicial objects the idea is that if you have a monad $T$ on a category $C$, then the bar construction turns each $T$-algebra into a simplicial $T$-algebra, using the comonad $F^T U^T$ on $C^T$ arising from $T$. If you apply $U^T$ to this simplicial algebra you get a simplicial object of $C$, and it turns out that the unit of $T$ supplies a contracting homotopy for this. (Compare the example at split coequalizer.)

   Mac Lane says that the complex $L^* a$ is a resolution, even though the associated simplicial object of $A$ won’t be contractible in general. Maybe it’s exact for some other reason, but you’ll have to ask someone who knows more homological algebra than I do (i.e. pretty much anyone).

5. • CommentRowNumber5.
   • CommentAuthorYaron
   • CommentTimeMay 19th 2012
   • PermaLink
   Author: Yaron
   Format: MarkdownItexFinn, thanks! I will try to work out the details of your explanation - this now looks as a manageable task. > pretty much anyone ...certainly not me :)

   Finn, thanks! I will try to work out the details of your explanation - this now looks as a manageable task.

   pretty much anyone

   …certainly not me :)