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I'm trying to take familiarity with homotopy theory and I have the following questions. Let $\mathcal{C}$ be a small category, and let $F\: : \:\mathcal{C}\to \mathcal{M}$ that take values in a (simplicial) model category. For any simplicial presheaves $K\: : \: \mathcal{C}^{op}\to sSet$, I denote with $$ K\otimes_{\mathcal{C}} F $$ the functor tensor product. Let * be the constant simplicial presheaf, it is easy to note that $$ *\otimes_{\mathcal{C}} F=colim F $$ Now choose a cofibrant replacement P of * in the projective model structure of simplicial presheaves. If we assume that $F$ is pointwise cofibrant, then the homotopy colimit is given by $$ P\otimes_{\mathcal{C}} F $$ A good candidate for the cofibrant replacement for $*$ is the nerve functor $N(-/ \mathcal{C})$. Another strategy to compute the homotopy colimit is given by the bar construction $B(K, \mathcal{C}, F)$ (see for example www.math.harvard.edu/~eriehl/hocolimits.pdf or the book categorical homotopy theory). In particular it is possible to show that there is a natural isomorphism $$ B(*, \mathcal{C}, F)\to N(-/ \mathcal{C})\otimes_{\mathcal{C}} F $$ Thus in some sense the bar construction contains the data of a cofibrant replacement for the simplicial presheaf $*$. Here are my questions:

1) Is it possible to generalize the above facts for a general simplicial presheaf $K\: : \: \mathcal{C}^{op}\to sSet$, i.e there exists a cofibrant replacament QK of K and a natural isomorphism $$ B(K, \mathcal{C}, F)\to QK\otimes_{\mathcal{C}} F? $$

2) If not, it is possible to find at least a weak equivalence?

Another formulation of this question is as follows: since $K\otimes_{\mathcal{C}} F$ may be intepretated as a weighted colimit, does the bar construction compute the homotopy weighted colimits?

3)(i don'think make sense) If $\mathcal{M}$ is not a simplicial category, it is possible to find a relation between the homotopy weighted colimit and the Bar construction?

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  • $\begingroup$ What is the $\mathcal{D}$ in the above bar construction? $\endgroup$ – Jonathan Beardsley Feb 18 '15 at 19:30
  • $\begingroup$ Ups is typo! I mean $\mathcal{C}$. I use the same notation of the book: categorical homotopy theory. $\endgroup$ – Cepu Feb 18 '15 at 19:35
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    $\begingroup$ Short answer to final question: yes, the bar construction computes homotopy weighted colimits. $\endgroup$ – Zhen Lin Feb 19 '15 at 11:49
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Yes. (Of course for these constructions to be homotopically well-behaved you need $F$ to be levelwise cofibrant.) In fact, assuming that such $QK$ exists we can express it by an explicit formula which then proves that it indeed exists. All we need to know is that tensoring over $\newcommand{\C}{\mathcal{C}}\C$ with a functor represented by $x \in \mathcal{C}$ is just evaluation at $x$. Hence $QK_x = QK \otimes_\C \C(x,-) = B(K, \C, \C(x,-))$ and that's how we construct $QK$.

Now the formula for $B(K, \C, F)$ holds for general $F$.

$$QK \otimes_\C F = B(K, \C, \C(-,-)) \otimes_\C F \\ = |\coprod_{y_0, \ldots, y_m} \C(-, y_0) \times \C(y_0, y_1) \times \ldots \times \C(y_{m-1}, y_m) \times K_{y_m}| \otimes_\C F \\ = |\coprod_{y_0, \ldots, y_m} F \otimes_\C \C(-, y_0) \times \C(y_0, y_1) \times \ldots \times \C(y_{m-1}, y_m) \times K_{y_m}| \\ = |\coprod_{y_0, \ldots, y_m} F y_0 \times \C(y_0, y_1) \times \ldots \times \C(y_{m-1}, y_m) \times K_{y_m}| = B(K, \C, F)$$

Moreover, $QK$ is projectively cofibrant. Indeed, we can explicitly write it as a cell complex in the projective model structure on diagrams $\C^{\mathrm{op}} \to \mathsf{sSet}$. For each $m$ there is a functor $G_m K \colon \C^{\mathrm{op}} \to \mathsf{Set}$ where $G_m K_x$ is the set

$$\{ (y, z) \mid y \colon [m] \to x \downarrow \C, z \in K_{y_m}, (y,z) \text{ is a non-deg. simplex of } N(x \downarrow \C) \times K_{y_m} \}$$

and a pushout square

$$ \begin{array}{ccccccccc} G_m K \times \partial\Delta[m] & \xrightarrow{} & \mathrm{Sk}^{m-1} QK & \newline \downarrow & & \downarrow & & \newline G_m K \times \Delta[m] & \xrightarrow[]{} & \mathrm{Sk}^m QK .& \end{array} $$

It remains to see that the map $QK \to K$ is a weak equivalence, but this follows from the fact that $\C(x,-)$ is projectively cofibrant since that map coincides with $K \otimes_\C B(\C(-,-),\C,\C(x,-)) \to K \otimes_\C \C(x,-)$ by a computation similar to the one above.

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  • $\begingroup$ The diagram doesn't display properly and I can't figure out why. I would be grateful if somebody could help me here. $\endgroup$ – Karol Szumiło Feb 18 '15 at 21:12
  • $\begingroup$ Thanks! I think that in this case we can remove form your formula for the bar construction the terms $\mathcal{C}(y_{i},y_{j})$, since $\mathcal{C}$ is not enriched (or trivial enriched simplicial category). $\endgroup$ – Cepu Feb 19 '15 at 9:54
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    $\begingroup$ They are discrete but not trivial so you cannot literally remove them. You can if you index coproducts over sequences of arrows instead of sequences of objects. It was just a notational choice. $\endgroup$ – Karol Szumiło Feb 19 '15 at 11:14
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    $\begingroup$ Uhm yes, you have right, it is not trivial is the simplicial structure:)! By the way it sems to me that the proof holds even in an arbitrary model category (you take a cosimplicial frame to get the tensor product). I say that because the functor tensor product between a cofibrant simplicial presheaf and a cofibrant diagram is independent by the choice of the frame. What do you think? $\endgroup$ – Cepu Feb 19 '15 at 11:44
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    $\begingroup$ The lack of strict enrichment is certainly an obstacle but you can work around that. It takes some effort so I'm not going to explain it here, but Hirschhorn discusses all the necessary techniques. $\endgroup$ – Karol Szumiło Feb 19 '15 at 20:05

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