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The left adjoint to the inclusion $sGrp\hookrightarrow sPtSet$ of the category of simplicial groups into the category of pointed simplicial sets is homotopy equivalent to the loop suspension functor $sGrp\leftarrow sPtSet:\Omega\Sigma$. This can be proved by observing that

  • looping and delooping $B:sGrp \rightleftarrows sSet_0:\Omega$ is a homotopy equivalence between reduced simplicial sets and simplicial groups.
  • Looping and suspension $\Sigma:sPtSet \rightleftarrows sSet_0:\Omega$ are adjoints.

The content of this is that Milnor's construction $F[X]$ of the reduced free simplicial group on a pointed simplicial set $X$ is of homotopy type $\Omega\Sigma X$.

Similarly, the left adjoint to the inclusion $sMon\hookrightarrow sPtSet$ from the category of simplicial monoids is also homotopy equivalent to $\Omega\Sigma$. The content of this is that the James construction $J(X)$ of the reduced free simplicial monoid on a pointed simplicial set $X$ is also of homotopy type $\Omega\Sigma X$.

I would like to know the left adjoints, up to homotopy, of the following inclusion functors:

  1. $sGrp\hookrightarrow sGrpd$ where $sGrpd$ is the category of simplicial groupoids
  2. $sMon\hookrightarrow sCat$ where $sMon$ is the category of simplicial monoids

This information would provide alternative model theoretic proofs of the corresponding algebraic constructions of simplicial objects, as with Milnor and James' constructions.

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    $\begingroup$ Won't this be the pointwise adjoints to inclusions of groups in groupoids (or monoids in categories)? If I'm not mistaken, the left adjoint to $Mon \to Cat$ is given by sending a category $\mathcal{C}$ to the monoid generated by morphisms in $\mathcal{C}$ and modulo relations coming from composition laws in $\mathcal{C}$. $\endgroup$ Nov 16, 2011 at 4:55
  • $\begingroup$ yeah, the algebraic constructions are easy to see. the part not obvious to me is how to find the homotopy type. $\endgroup$
    – user2529
    Nov 16, 2011 at 8:53
  • $\begingroup$ @ColinTan the homotopy they of what? $\endgroup$ Feb 15, 2015 at 22:37

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