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Let $\mathcal A, \mathcal A', \mathcal B$ be dg-categories over a field $k$ (this assumption allows me not to derive the tensor product, I don't think it is really essential). Let $F : \mathcal A \to \mathcal A'$ be a dg-functor. This clearly induces a dg-functor $F' :=F \otimes 1_{\mathcal B} : \mathcal A \otimes \mathcal B \to \mathcal A' \otimes \mathcal B$, which is fully faithful if $F$ is such.

A $\mathcal A$-$\mathcal B$-bimodule $G$ ($\mathcal A$ acting on the left, $\mathcal B$ acting on the right) is by definition a right $\mathcal A^{\text{op}} \otimes \mathcal B$-dg-module, that is, a dg-functor $\mathcal A^{\text{op}} \otimes \mathcal B \to \mathbf{C}_{\text{dg}}(k)$, where $\mathbf{C}_{\text{dg}}(k)$ is the dg-category of cochain complexes of $k$-modules. It is well-known (see this, for example) that dg-modules can be extended along dg-functors. My question is the following. Let $G$ be a $\mathcal A$-$\mathcal B$-bimodule, and let $\mathrm{Ind}_{F'}(G)$ its extension along $F'$. Now, assume that $G$ is a quasi-functor, that is, $G(A,-)$ is quasi-isomorphic to a representable right $\mathcal B$-dg-module for all $A \in \mathcal A$. Then, is $\mathrm{Ind}_{F'}(G)$ again a quasi-functor? If needed, assume that $F$ is fully faithful.

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  • $\begingroup$ What do you mean by fully faithful in this context? Homotopically or strictly? $\endgroup$ – Fernando Muro Oct 10 '14 at 15:04
  • $\begingroup$ I assume "strictly". It is already interesting in many situations. $\endgroup$ – Francesco Genovese Oct 10 '14 at 16:11
  • $\begingroup$ Hi Francesco, can you be more precise sketching the extension of a bimodule along a functor? Seems like you refer to an "extension" of a $\cal V$-profunctor along a $\cal V$-functor, but I haven't heard anything about it before $\endgroup$ – Fosco Oct 14 '14 at 8:46
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Once that you clarified me the meaning of $\text{Ind}_{F'}(\Gamma)$ as an "extension", I think I have half of the answer: what you are looking for is the fact that composition of representable profunctors is again a representable profunctor: some coend juggling shows that $$\text{Ind}_{F'}(\Gamma) := \text{Lan}_{F\otimes 1}\Gamma\cong \hom(F,1)\diamond \Gamma$$ (where $\diamond$ is the composition of profunctors), and hence if we let $\Gamma=\hom(G,1)$ we get $\text{Ind}_{F'}(\Gamma) \cong\hom(FG,1)$; same story if we compose $\hom(1,F)$ with $\hom(1,G)$.

here's (part of) the coend juggling: $$\begin{align*} \text{Ind}_{F'}\Gamma(c',d') &\cong \displaystyle \int^{c,d}\mathcal{C}'(F^\text{op}c,c')\cdot\big(\mathcal{D}(d,d')\cdot G(c,d)\big) \\ &\cong \displaystyle \int^c \mathcal{C}'(F^\text{op}c,c')\cdot\left(\int^d \mathcal{D}(d,d')\cdot G(c,d)\right)\\ &\cong \displaystyle \int^c \mathcal{C}'(F^\text{op}c,c')\cdot G(c,d')\\ & \cong \text{Lan}_{F^\text{op}}(G(-,d'))(c') \end{align*}$$ now compare this with the rule giving you the composition of profunctors (it's important to have $F^\text{op}$!).

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  • $\begingroup$ This should actually answer my question, since the composition of quasi-functors is again a quasi-functor. One probably has to derive something, but it should work just fine. $\endgroup$ – Francesco Genovese Oct 14 '14 at 10:04

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