# topological monoid from symmetric monoidal category

What is the standard reference for the fact that the classifying space of a strict monoidal category is a topological monoid with respect to the operation induced by the tensor product?

EDIT: The first version of the question was stated for strict symmetric monoidal categories, but as was pointed out in the comments, a symmetry is of course not necessary to just get a monoid structure on the classifying space.

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Of course, you can drop the word "symmetric": a strict monoidal category is the same as a monoid in $(Cat, \times)$, and since the classifying space functor $Cat \to kSpace$ preserves products, it takes monoids to monoids. This isn't covered in Segal's Categories and Cohomology Theories? (Maybe it is; I don't recall.) – Todd Trimble Nov 27 '12 at 19:26
@Todd: Thank you. I will have a look. I know that the proof of this is really just the fact that geometric realizations are product preserving. But I happen to write a paper, which uses this and has a main audience outside of topology. So I need to back things up with as many references as possible. – Ulrich Pennig Nov 27 '12 at 19:49
Theo, if you're asking Todd, then he'd say "topological monoid" here is an abuse of language where strictly speaking we are taking $kSpace$ as our "convenient" category of topological spaces. I thought that was a pretty standard maneuver; it's well known that $Top$ has some undesirable properties (such as not being cartesian closed, when we'd really like function spaces with all our hearts). – Todd Trimble Nov 27 '12 at 23:12
As Todd already pointed out, this is special case of the observation that every lax monoidal functor $C \to D$ extends to a functor $\mathsf{Mon}(C) \to \mathsf{Mon}(D)$ (applied to $C=(\mathsf{Cat},\times)$ and $D=(\mathsf{CGHaus},\times)$). Since this is trivial, it is always just mentioned in the literature (for example Saavedra Rivano, Categories Tannakiennes, I.6.1.4.). – Martin Brandenburg Nov 28 '12 at 0:25
You not only can but should drop the word symmetric''. Otherwise someone in a naive audience may ask whether your topological monoid is commutative, and of course it is not. With the standard notion of a strict symmetric monoidal category (aka a permutative category), the classifying space gives rise to a spectrum whose zeroth space is a group completion of your monoid. – Peter May Dec 2 '12 at 23:47

Well, if you are going to reference me somewhere, I can give you something more explicit. The cited Corollary 11.7 is only about topological monoids. However Theorem 4.10 of "$E_{\infty}$ spaces, group completions, and permutative categories" ( http://www.math.uchicago.edu/~may/PAPERS/13.pdf ) has the precise statement requested: "If $(\mathcal{A},\Box,\ast)$ is a strict monoidal category, then $B\mathcal{A}$ is a topological monoid with product $B\Box$." The result goes on to say precisely what holds with respect to commutativity when $\mathcal{A}$ is permutative (= strict symmetric monoidal).