TITLE: homotopy tensor product of functors and bar construction
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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.