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?

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    $\begingroup$ 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. $\endgroup$ Mar 4 at 15:49
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    $\begingroup$ 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. $\endgroup$
    – IJL
    Mar 4 at 16:46
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    $\begingroup$ 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. $\endgroup$ Mar 5 at 3:27

1 Answer 1


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 $|(BG)^+| \simeq (|BG|)^+$.

  • $\begingroup$ 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. $\endgroup$ Mar 4 at 17:18
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    $\begingroup$ 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/… $\endgroup$ Mar 4 at 17:20
  • $\begingroup$ 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? $\endgroup$
    – wind
    Mar 5 at 4:33

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