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Example 11.4 in this paper claims that the tensor product of chain complexes of bimodules (over not-necessarily-commutative rings) satisfies the pushout-product axiom (the first condition of a Quillen bifunctor). That is, if $f : A \to B$ and $g : X \to Y$ are cofibrations (in the projective model structure on chain complexes), then the map $f \square g : (A \otimes Y) \coprod_{A \otimes X} (B \otimes X) \to B \otimes Y$ induced by $f \otimes 1 : A \otimes Y \to B \otimes Y$ and $1 \otimes g : B \otimes X \to B \otimes Y$ is also a cofibration. My question is: why does the tensor product satisfy the pushout-product axiom?

Let $f : A \to B$ be a map of chain complexes of $R$-$S$ bimodules which is levelwise injective with projective cokernel (that is, a cofibration), and let $g : X \to Y$ be a cofibration between complexes of $S$-$T$ bimodules. If I'm not mistaken, $f \square g$ is levelwise injective, and $$\mathrm{coker}(f \square g)_k \cong \bigoplus_{p+q=k} \mathrm{coker}(f_k) \otimes_S \mathrm{coker}(g_k).$$ By assumption, $\mathrm{coker}(f_k)$ is a projective $R$-$S$ bimodule, and $\mathrm{coker}(g_k)$ is a projective $S$-$T$ bimodule. But I don't think it follows that their tensor product is a projective $R$-$T$ bimodule. (It would if $\mathrm{coker}(g_k)$ was instead projective as a right $T$-module.)

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