# When do dg-lifts exist?

Let $\mathcal A$ and $\mathcal B$ be abelian categories (with enough injectives and countable products) and $$F:D^+(\cal A) \rightarrow D^+(\cal B)$$ be a triangulated functor. I am interested whether there exists a dg-lift, by which I mean dg-functor $$\tilde F:C^+(inj \cal A)\rightarrow C^+( inj \cal B)$$ between the dg-categories of bounded below complexes with injective entries and usual $Hom^\bullet$-complexes as morphisms, which induces $F$ on homotopy categories. I am also interested in variants with $D^+$ replaced by some other nice subcategory of the derived category, projectives instead of injectives etc.

Now I have heard from time to time the statement, that

"Any functor encountered in practice has a dg- lift. "

Yet I am not really aware of criteria which guarantee dg-lifts, I only know one recipe to construct dg-lifts:

First step: Try to lift F to a dg-functor $$C^+(inj \cal A)\rightarrow C^+(\cal B)$$ This step is often but not always obvious. For example it works when $F$ is a derived functor. And it is not obvious to me for example if $\cal A$ is the heart of a t-structure on $D^b(\cal B)$ and $F$ is the realization functor: $real:D^b(\cal A)\rightarrow D^b(\cal B)$.

Second step: Hope that injective resolutions can be choosen in a nice way, giving a dg-functor $$C^+(\cal B)\rightarrow C^+(inj \cal B)$$ and compose this functor with the one from step 1. This should for example work if $\cal B$ is the category of modules over a sheaf of k-algebras (where k is a field) and maybe doesn't if $\cal B$ is the category of abelian groups.

Are there other methods/criteria that allow to construct dg-lifts? I m especially interested in situations where step 1 is not obvious.

What are examples of such functors $F$ which do not have dg-lifts, even if $\cal A$ and $\cal B$ have enough injectives/projectives? How does one proof such a statement?

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The question, stated in terms of abelian categories, is more subtle, but you could extend your question to differential graded algebras, i.e. instead of the derived category of an abelian category such as the category of modules over a ring, consider the derived category of a differential graded algebra. In tha case the following paper provides with a very explicit example where an equivalence of derived categories cannot be dg-lifted

Dugger, Daniel(1-OR); Shipley, Brooke(1-ILCC-MS) A curious example of triangulated-equivalent model categories which are not Quillen equivalent. (English summary) Algebr. Geom. Topol. 9 (2009), no. 1, 135–166.

The categories are the derived categories of dg-algebras $A$ and $B$, where $A=\mathbb{F}_p[x^{\pm 1}]$, with $x$ in degree $1$ and trivial differential $d=0$, and $B=\mathbf{Z}\langle x^{\pm1},e\rangle/(e^2,ex+xe-x^2)$ with $e$ and $x$ in degrees $2$ and $1$, respectively, and differential $d(e)=px$ and $d(x)=0$. Here $p\in\mathbb{Z}$ is a prime number. Notice that $A$ and $B$ are not commutative, except from $A$ when $p=2$, and $A$ is always the homology of $B$. Both dg-algebras have the same defive category: the category of $\mathbb{F}_p$-vector spaces with the identity as the translation functor and $3$-periodic exact sequences as exact triangles.

A general answer to your question is complicated and has to do with $A_\infty$-obstruction theory.

If you wish to enlarge even more the kind of derived categories to be considered and allow derived categories of ring spectra, the following paper yields a nice example of non-realizability of a functor which is not an equivalence

Neeman, Amnon(1-VA) Stable homotopy as a triangulated functor. Invent. Math. 109 (1992), no. 1, 17–40.

On the one hand you have the derived category of $\mathbb{Z}[\frac{1}{2}]$ and on the other hand the derived category of $S[\frac{1}{2}]$, where $S$ is the sphere spectrum.

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