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I have been reading the paper by Toën "The homotopy theory of dg categories and derived Morita theory" where in chapter 4 it is stated that the tensor product of two cofibrant dg categories $C$ and $D$ (defined by $obj(C\otimes D)=objC\times obj D$ and with space of morphisms obtained by tensoring the two chain complexes in $C$ and $D$) is not in general cofibrant, and this is a problem in both the "standard" and the Morita model structures on the category $\mathbf{dgCat}_k$ (here $k$ is a commutative ring with 1 and with arbitrary characteristic).

It is also known (proposition 2.3 (2) in the same article) that a cofibrant replacement functor can be chosen in such a way that it is the identity on objects, therefore the problem seems to lie exclusively in what happens to morphisms. The same proposition contains the result that a cofibrant dg category has morphisms spaces which are cofibrant complexes wrt the projective model structure on $\mathbf{Ch}(k)$.

A cofibrant chain complex $X_\bullet$ has $X_n$ projective $\forall n$ and the converse is true only if $X_\bullet$ is bounded below. Nevertheless, the tensor product (and the direct sum) of two projective modules is again projective so I think I am missing some essential point: given $X_\bullet$ and $Y_\bullet$ cofibrant chain complexes, $(X\otimes Y)_\bullet$ will be made of projective modules, so I think I need to find an unbounded cofibrant complex which tensored with some other cofibrant complex loses cofibrancy.

This result was cited in several other places, but I was unable to find even a discussion longer than the statement. I would be glad if someone could give me such an example, or point out some reference where I can find it.

Thank you in advance

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You can consider DGAs, which are DG-categories with only one object. The polynomial ring $k[x]$ concentrated in degree $0$ (with trivial differential) is cofibrant, since it is free as a (graded) $k$-algebra. Take another copy, $k[y]$. The tensor product $k[x]\otimes k[y]=k[x,y]$ is not cofibrant as a DGA. Indeed, you can construct a cofibrant resolution $C\stackrel{\sim}\twoheadrightarrow k[x,y]$ such that $C_0=k\langle x,y\rangle$ is a free non-commutative $k$-algebra on the same two generators, and if $k[x,y]$ were cofibrant, it would be a retract of $C_0=k\langle x,y\rangle$ by the lifting axiom, but that's impossible because $k[x,y]$ should then also be free as a non-commutative $k$-algebra.

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  • $\begingroup$ thank you! I had myself convinced that things would have to be quite pathological for cofibrancy to fail, that's life. $\endgroup$ – James Feb 11 '15 at 19:35
  • $\begingroup$ Nice example! One could say that the "monoid axiom", which holds for chain complexes, guarantees that the monoids get a model category structure, but it doesn't mean that the category of monoids itself has to be a monoidal model category (for monoidal model categories, the tensor product of two cofibrant objects is cofibrant). $\endgroup$ – Bruno Stonek Mar 3 '17 at 16:01
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Let $\Delta^1_k$ be the $k$-linear dg-category with two objects $0$ and $1$, mapping complexes $$ Map(0,0) = [k], $$ $$ Map(0,1) = [k], $$ $$ Map(1,0) = [0], $$ $$ Map(1,1) = [k] $$ where $[k]$ means the complex with $k$ in degree zero, and a composition law defined in the obvious way. This is a cofibrant dg-category, and one can show that the tensor product $$ \Delta^1_k \otimes \Delta^1_k $$ is not cofibrant. For an argument, see Exercise 3.2.2 in [Bertrand Toën, Lectures on DG-categories, in "Topics in algebraic and topological K-theory", pp. 243-302 (2011)], available here (labelled as Exercise 14 in the latter pdf).

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  • $\begingroup$ thank you! I worked out the exercise and I feel more grown up now. I knew thos awesome lectures by Toen, but somehow didn't pay attention to those exercises $\endgroup$ – James Feb 11 '15 at 19:28

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