# Why does tensor product satisfy pushout-product axiom?

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.)