# Gamma spaces and monoidal categories II

This question is kind of a follow-up of this one.

Suppose I have a topological category $\mathcal{C}$ (objects and morphisms topological spaces, source and target map continuous, etc.) together with a continuous tensor product $\otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}$, such that it is strict monoidal and symmetric, but there is no unit object (I have some kind of homotopy unit, in the cases I am interested in, but I don't know how to build it into the category).

What is the structure of the classifying space $B\mathcal{C}$? Does the Gamma-space construction of Segal still work and give me some kind of homotopy associative $H$-space?

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I think you do not need to use the Gamma space construction of Segal here. You just need to note that the classyfing space functor from topological categories into toopological spaces preserves products (given you work with compactly generated spaces).

This implies that in your situation the classifying spaces $|C|$ inhertits a strict multiplication, i.e. it is a topological monoid (possibly without unit). The homotopy unit on the topoloigal category will then lead to a homotopy unit for your monoid.

If you now want to group complete you form $\Omega B |C|$. Note here that $B|C|$ can be formed by taking the fat geometric realization of the simplicial space $N|C|_n := |C|^{n-1}$. For this you don't need degeneracies, i.e. units.

Or do I misunderstand something?

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You are right. But this way, I only see the $H$-space structure of the geometric realization. In a Gamma-space I also have control over the higher homotopies. For example, I can deloop a Gamma space. The question is, whether I still get a Gamma space, if I drop the assumption about units. –  Ulrich Pennig Mar 1 '12 at 9:26
You don't see higher homotopies because they are not there. You can deloop topological monoids without units, as a described above. If you insist on getting a Gamma space I don't know from the top of my head. I have to think about it. –  Thomas Nikolaus Mar 1 '12 at 9:30
There are higher homotopies still; they come from the symmetric structure on the classifying space, and are required in order to allow iterated delooping. –  Tyler Lawson Mar 1 '12 at 14:51
I understood that Ulrich assumed that the initial category was strict symmetric!? If this is not the case you are of course right (and in fact it almost never happens that its strict symmetric). –  Thomas Nikolaus Mar 1 '12 at 15:36
@Tyler: So, iterated deloopings still work in the non-unital case? –  Ulrich Pennig Mar 1 '12 at 21:15

You can construct a $\Gamma_{epi}$-category from your category (where $\Gamma_{epi}$ denotes the category of finite pointed sets with epimorphisms). You cannot extend it to a $\Gamma$-category if your symmetric monoidal category doesn't have a unit (if you could, this would give you a unit). When you apply nerve you get a $\Gamma_{epi}$-space and this is probably the same as a non unital $E_{\infty}$-space (i.e. a space over the operad $E_{\infty}$ where you forgot about the $0$-th space).

However there is a result of Lurie in Higher Algebra that says that homotopy units on non unital $E_{\infty}$-spaces can be strictified so that the result is a unital $E_{\infty}$-space. You seem to have a homotopy unit in your category so this result might be relevant.

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Thank you. That makes sense. I also wondered about the following thing: I could just add a dummy object 1 to the category with mor(1,x) = empty if x is not 1 and mor(1,1) = id_1. Is there any problem with extending the monoidal structure in such a way that 1 becomes a unit? If this works: What is the relation between the delooping above and the delooping of the "unification"? –  Ulrich Pennig Mar 2 '12 at 20:18
What you do is adding a disjoint unit or equivalently writing you non unital monoid as the augmentation kernel of an augmented monoid. I can tell you what happens for commutative algebras in the category of chain complexes. I don't know how relevant it is to this situation. So if A is an augmented algebra and I is its augmentation ideal, then the iterated bar $B^n(A)$ is equivalent to $k\oplus B^n(I)[n]$. You can find this result in a paper of Po Hu. –  Geoffroy Horel Mar 3 '12 at 2:31