MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Let $C$ be an $(\infty,1)$-topos. The $(\infty,1)$-category of group objects in $C$ is a full sub-$(\infty,1)$-category of groupoid objects in $C$: $${\mathsf{Grp}}(C) \hookrightarrow {\mathsf{Grpd}}(C)$$

Is this full subcategory reflective? Here is one way to go about constructing a $(\infty,1)$-functor in the other direction: $${\mathsf{Grpd}}(C)\xrightarrow{{\mathbf{B}}} C \xrightarrow{R} {\mathsf{PointedConnected}}(C)\xrightarrow{\Omega} {\mathsf{Grp}}(C)$$ The functor $R: C\to {\mathsf{PointedConnected}}(C)$ takes an ordinary object $X$ in $C$ and gives its pointed-connected-reflection, given, say, by taking the homotopy cofiber of the inclusion from the 0-truncation $X_0$ to $X$. Conjecturally, the above $(\infty,1)$-functor could be the group reflection of a groupoid object, but it's really a guess.

So, the question: is ${\mathsf{Grp}}(C)$ reflective in ${\mathsf{Grpd}}(C)$? How can we give the adjoint?

share|cite|improve this question
How do you get an inclusion from the $0$-truncation $X_0$ to $X$? – Charles Rezk Aug 28 '13 at 13:36
At least, if $X=BA$ is the delooping of a groupoid object $A$, then $X_0=BA_0$. In this case, the morphism $X_0\to X$ is induced by $A_0\hookrightarrow A$ which includes into $A$ the discretization of $A$. – user2529 Aug 29 '13 at 2:54
up vote 10 down vote accepted

If $\mathcal{C}$ is an $\infty$-topos, the $\infty$-category of groupoid objects of $\mathcal{C}$ is equivalent to the full subcategory of $Fun( \Delta^1, \mathcal{C})$ spanned by the effective epimorphisms $X \rightarrow Y$. Under this equivalence, the group objects correspond to the full subcategory where $X$ is a final object. The inclusion has a left adjoint, which carries an effective epimorphism $f: X \rightarrow Y$ to the induced map $\mathbf{1} \rightarrow Y \amalg_{X} \mathbf{1}$.

share|cite|improve this answer

Your Answer


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