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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, $\bar{X}\simeq \Omega\Sigma X$.

edited Nov 5 2011 at 5:52

asked Nov 4 2011 at 14:39

Colin Tan
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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 $\Omega\Sigma X$. – Moosbrugger Nov 4 2011 at 15:14

I think "pointed monoidal" is redundant -- isn't every monoidal category uniquely pointed by its unit object? – Mike Shulman Nov 4 2011 at 23:35

Moosbrugger, does "computed by" mean "homotopic to"? – Colin Tan Nov 5 2011 at 5:51

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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? – Dylan Wilson Nov 5 2011 at 21:00

dagness · Answer 1 · 2011-11-08 08:37:52Z UTC

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The isomorphism fS0B = S0A implies the equivalence fS0B ' S0A since S0B is qf-fibrant.

answered Nov 8 2011 at 8:37

dagness
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A kind of James construction for $\infty$-groupoids

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