Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Every simplicial set is the colimit of its finite simplicial subsets. Suppose $G$ be a finite discrete set.

Is every simplicial $G$-set a colimit of its finite simplicial $G$-subsets? I'm particularly interested in the case where $G={\mathbb{Z}}/2$.

share|cite|improve this question
"Every simplicial set is the colimit of its finite simplicial subsets. " Really? I doubt that this is true. Or what notion of "finite" are you working with here? – Martin Brandenburg Oct 25 '12 at 15:15
The obvious way of making the statement true is to replace "finite" by "finitely presented" (in the sense of Gabriel and Ulmer). But I think this turns out to be the same thing as a simplicial set with finitely many non-degenerate simplices. – Zhen Lin Oct 25 '12 at 16:29

I assume you meant $G$ should be a finite discrete group? Also, I think the answer is yes. Every simplex of a $G$-sset is contained in a finite sub-sset, and then you you can take the closure under the $G$-action (which will still be finite, since $G$ is). This closure will be a sub-sset because the face and degeneracy maps are $G$-equivariant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.