# Comparison of products in Quillen and Waldhausen K-theory

I'm relatively new to algebraic K-theory and stumbled upon the following question. I would be very glad If someone could provide a reference to an answer or a short argument.

We are given an exact category $\mathcal{E}$ with a compatible tensor product $\otimes$, i.e. a bifunctor $\otimes:\mathcal{E} \times \mathcal{E} \to \mathcal{E}$ that is exact in both variables.

In his paper "Algebraic K-theory of Generalized Free Products, Part 1", Waldhausen uses the double $Q$ construction to obtain an induced product in Quillen K-theory

$\mathrm{K}_i(\mathcal{E}) \times \mathrm{K}_j(\mathcal{E}) \to \mathrm{K}_{i+j}(\mathcal{E})$.

But we can also consider $\mathcal{E}$ as a category with cofibrations (the admissible monomorphisms) and weak equivalences (the isomorphisms) and Waldhausen proves in "Algebraic K-theory of spaces" that the associated K-groups are isomorphic to Quillen's. In the same paper, he describes how a bifunctor like $\otimes$ (satisfying a further technical condition) induces a product similar to the above one on the Waldhausen K-groups of $\mathcal{E}$, by applying the $S_{\bullet}$ construction twice.

Is this the same product? That is, are the Quillen and Waldhausen K-theory of $\mathcal{E}$ isomorphic as rings?

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I'd say this is the kind of thing you want to give for granted without further checking ;-) –  Fernando Muro Dec 16 '13 at 18:52
Or you can look at (1) C. A. Weibel, "A survey of products in algebraic $K$-theory", Algebraic $K$-theory, Evanston 1980 (Proc. Conf., Northwestern Univ., Evanston, Ill., 1980), pp. 494–517, Lecture Notes in Math., 854, Springer, Berlin, 1981 and (2) Kazuhisa Shimakawa, "Uniqueness of products in higher algebraic K-theory", Publ. Res. Inst. Math. Sci. 24 (1988), no. 1, 99–120. –  John Rognes May 16 at 20:13