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Start with a pointed $\infty$-groupoid $X$. Consider $X$ as a pointed $\infty$-category and construct the free pointed monoidal $\infty$-category $\bar{X}$. This free construction $\bar{X}$ is at least an $\infty$-groupoid too. How is the homotopy type of $\bar{X}$ related to the homotopy type of $X$?

This seems to be a James construction in the context of $\infty$-groupoids, but I can't be sure. That is, conjecturally, $\bar{X}\simeq \Omega\Sigma X$.

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    $\begingroup$ I think your formula isn't quite right: consider the case $X=S^0$. The problem is that $\Omega\Sigma X$ is group-like while $\overline{X}$ is not. But it's direct to show that the group completion of $\overline{X}$ is indeed computed by $\Omega\Sigma X$. $\endgroup$
    – Moosbrugger
    Nov 4, 2011 at 15:14
  • $\begingroup$ I think "pointed monoidal" is redundant -- isn't every monoidal category uniquely pointed by its unit object? $\endgroup$
    – Mike Shulman
    Nov 4, 2011 at 23:35
  • $\begingroup$ Moosbrugger, does "computed by" mean "homotopic to"? $\endgroup$
    – user2529
    Nov 5, 2011 at 5:51
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    $\begingroup$ The theory of infinity-groupoids should be equivalent to the theory of spaces- they're just Kan complexes. So shouldn't the James construction for infinity-groupoids just be... the usual James construction for spaces? $\endgroup$
    – Dylan Wilson
    Nov 5, 2011 at 21:00

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