# Plus construction on Simplicial Sets?

I had asked this question in Math StackExchange a few days ago, but didn't get any answers. I believe its more suitable to be asked here.

Write $$\mathsf{sSet}$$ for the category of simplicial sets and $$\mathsf{Top}$$ for the category of topological spaces. I would like to know if there a functor $$\mathsf{sSet}\to\mathsf{sSet}$$ that resembles the plus construction in $$\mathsf{Top}$$?

More precisely, let $$G$$ be a (perfect) group and $$|BG|$$ the classifying space of $$G$$, by the geometric realization of the nerve construction (so $$BG$$ is the simplicial set with $$B_{n}G$$ = set of all tuples $$(g_1, \cdots ,g_n)$$, with the natural face and degeneracy maps. And, $$|BG| =$$ geometric realization of the simplicial set $$BG$$. I'm using the construction given in Weibel's "An introduction to Homological Algebra" book, Page 257).

The question is as follows :

Does there exist a functor (denoted with slight abuse of notation) $$(-)^{+}\colon\mathsf{sSet}\to\mathsf{sSet}$$, such that $$|(BG)^{+}| \cong (|BG|)^{+}$$ for any (perfect) group $$G$$?

In other words, can we make a plus construction in simplicial sets, so that it commutes with the geometric realization functor?

• This is not exactly an answer to your question, but might still be relevant to you. The paper arxiv.org/pdf/1711.08898.pdf proves an $\infty$-categorical universal property of the plus construction. This means that on the $\infty$-category of simplicial sets, the plus construction refines to a functor. You're of course asking for more, namely whether one can strictify this to a functor on the $1$-category of simplicial sets. Mar 4 at 15:49
• I am unsure whether this is correct, so I didn't want to put it as an answer. However, I think that the functor that sends a simplicial set to the simplicial abelian group whose $n$-simplices are the free abelian group with basis the $n$-simplices of the original simplicial set does this job.
– IJL
Mar 4 at 16:46
• No, the free abelian group is much more brutal, and produces a generalized Eilenberg-MacLane space whose homotopy groups are the homology groups of the original simplicial set. This is very different from the plus construction, which for example leaves simply connected spaces unchanged. Mar 5 at 3:27

The answer is yes. This is spelled out in the book The local structure of algebraic K-theory by Bjørn Ian Dundas, Thomas G. Goodwillie and Randy McCarthy. Check out Section 1.6.1 on page 26, where they explain the plus construction as a functor $$(-)^+: sSet\to sSet$$. They refer to page 219 of the following book:

A. K. Bousfield and D. M. Kan. Homotopy limits, completions and localizations. Springer-Verlag, Berlin, 1972. Lecture Notes in Mathematics, Vol. 304.

On page 219, Bousfield and Kan begin with a ring $$R$$ and define the partial $$R$$-completion $$C^R(X) = R_\infty Sing |X|$$ of a simplicial set $$X$$. The notation $$R_\infty X$$ is defined on page 41. Dundas, Goodwillie, and McCarthy use this for $$R = \mathbb{Z}$$, and define $$X^+ = C^{\mathbb{Z}}(X)$$.

Their Proposition 1.6.4 on page 27 proves that this $$X^+$$ satisfies the properties you would expect from Quillen's plus construction. Lastly, Theorem 1.6.5 proves that these properties characterize $$X^+$$ up to homotopy. That theorem is proven on page 255, in Appendix A of the book. It follows from that result that $$|(BG)^+| \simeq (|BG|)^+$$.

• Also relevant is page 191 of the Goerss-Jardine book: "Rather a lot of standard homotopy theory is amenable to proof by simplicial techniques. The reader may find it of particular interest to recast the Haussman-Husemoller treatment of acylic spaces and the Quillen plus construction [41] in this setting." They go on to sketch how to do the plus construction within sSet, but the Dundas et al book goes in more depth. Mar 4 at 17:18
• In the comments of the MSE version of the question, the OP says it's ok to have a canonical homotopy equivalence instead of a homemorphism, $|(BG)^+|\simeq (|BG|)^+$: math.stackexchange.com/questions/4873859/… Mar 4 at 17:20
• Wow, thank you so much! That's a very interesting (and complicated construction). By the way, how does the last conclusion $|(BG)^+| \simeq (|BG|)^+$ follow, from Theorem 1.6.5?
– wind
Mar 5 at 4:33