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A groupoid object in an $(\infty,1)$ category $\mathcal{C}$ is a functor $G:N(\Delta)^{op} \to \mathcal{C}$ such that for any partition $[n]=S \cup S'$ intersecting in $s$, the object $G([n]$ is the pullback of $G(S)$ and $G(S')$ over $G(s) \simeq G([0])$. In particular, $G$ is a group object if $G([0])$ is final.

Can anyone explain to me why this definition agrees intuitively with our notion of group objects in ordinary categories? For example, where do "inverses" come from? Can we consider all group objects as monoids in some monoidal category (if the monoid structure comes from products), i.e. is the functor $G$ lax monoidal?

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    $\begingroup$ I think you mean $\Delta$ instead of finite sets, and 'contractible' maybe means 'final object'. Use the decomposition $[2] = \{0,1\} \cup \{0,2\}$ and then lift the pair $(x,e)$ to something in $G^{\times 2}$, $(x,y)$. By definition, $xy = e$, and there's your inverse. $\endgroup$
    – Dylan Wilson
    Commented Feb 24, 2018 at 0:00
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    $\begingroup$ (If you're familiar with the 'Kan complexes are $\infty$-groupoids' point of view, then this `decomposition--> pullback' condition is equivalent to having fillers for all horns. See, e.g. HTT.6.1.2.6) $\endgroup$
    – Dylan Wilson
    Commented Feb 24, 2018 at 0:02
  • $\begingroup$ @DylanWilson Yes I do mean $\Delta$ and 'final object,' thanks. That makes sense. Is it easy to see as well that the functor is lax monoidal, if the monoidal structure is given by products? $\endgroup$
    – Exit path
    Commented Feb 24, 2018 at 0:02
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    $\begingroup$ The functor is monoidal (not just lax monoidal) for the ‘concatenate’ monoidal structure on the source. But it is also more than that because we’re allowed to decompose in more general ways $\endgroup$
    – Dylan Wilson
    Commented Feb 24, 2018 at 14:26

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