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I'm studying a proof of the fact that the category of dg-categories admits a (Dwyer-Kan) model structure. As references, I'm using Pieter Belmans' master thesis and Goncalo Tabuada's paper Une structure de catégorie de modèles de Quillen sur la catégorie des dg-catégories (where the proof has first appeared).

An important ingredient of the proof is the Lemma B.2 in Belmans' notes. I won't reproduce everything here, only the relevant part. Suppose we're given homogeneous morphisms $f\colon X\to X$ and $g\colon X\to Y$ in some dg category $C$, of degrees $-1$ and $0$, respectively. Let $Y_C\colon C\to \mathrm{dgMod}\text{-}C$ be the embedding of $C$ into the dg-category of right $C$-modules, i.e. dg-functors $C^{\mathrm{op}}\to C_{\mathrm{dg}}(k)$. Unless I'm misunderstanding him, Belmans essentially claims that "because of Koszul sign rule" we have $Y_C(g\circ f) = -Y_C(g)\circ Y_C(f)$. Tabuada doesn't treat this part in detail, but for his proof of the lemma to work, the identity should be satisfied as well. On the other hand, if $f'\colon Y\to Y$ is also of degree $0$, we should have $Y_C(f'\circ g) = Y_C(f')\circ Y_C(g)$ (according to the proof).

Now $Y_C(h)$, for homogeneous $h$ of degree $i$, should give a dg-natural transformation $\mathrm{Hom}(-,h)$. But what is the sign convention for this? The result is well-known so I doubt there is a mistake and only blame my ignorance. Still, a possibility is open, and for that reason I provide context.

Edit: It appears an (only?) reasonable sign convention for this is as follows: $\mathrm{Hom}(-,X)(f)(g) = (-1)^{\deg(f)\deg(g)} g\circ f$ and $\mathrm{Hom}(-,f)_{X'}(g) = f\circ g$. Otherewise either $\mathrm{Hom}(-,f)$ is a not a dg natural transformation of degree $\deg(f)$, or $Y_C$ doesn't commute with differentials. This raises the question, however: is there a mistake in the proof? If there is, I'm sure it can easily be rectified since the result is widely accepted. Of course, it's more likely that I'm missing something, and would be immensely grateful if someone could point out what exactly is I'm missing here.

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    $\begingroup$ It may be the case that there is more than one typo in the proof of Belmans. For example, take a look at the formula (B.7) in his proof: one can expect the diagonal entries r_2' and r_1' to be endomorphisms of Y respectively X, yet Belmans claims they are morphisms Y→X and X→Y respectively. Something seems to be amiss… $\endgroup$
    – Dmitri Pavlov
    Feb 12, 2022 at 23:47
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    $\begingroup$ @DmitriPavlov Yes, there are typos, but they are either inconsequential or easily corrected comparing with Tabuada (as far as I've studied the proof further). As for this one, however, your comment has led me to try out the computation again and I had a silly idea - take $r'_2$ (not $r'_1$) and $r'_{1,2}$ to be not the components of $h$ per se, but their additive inverses. This way, we still have a bijection, and later theorems about quasi-equivalences should still go through, I think, since cohomology doesn't see signs. $\endgroup$
    – Jxt921
    Feb 13, 2022 at 7:36

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