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Sep 27, 2019 at 13:49 comment added Simon Henry @Nicky : One first check that $\pi_0: \widehat{\Delta} \rightarrow Set $ is left adjoint to the constant simplicial set functor. Then one deduce that applying level wise $\pi_0$ : $sPre(C) \rightarrow Pre(C)$ is left adjoint to the inclusion $ Pre(C) \rightarrow sPre(C)$. Then one uses that sheaficication is left adjoint to the forgetful inclusion of sehaves into presheaves, and that adjoint compose.
Sep 6, 2019 at 17:10 comment added Nicky Thanks! Can you explain a bit why this is true in 1-category level please?
Sep 6, 2019 at 14:01 vote accept Nicky
Sep 6, 2019 at 11:31 comment added Simon Henry Yes it is. Both at 1-category level and homotopicaly speaking.
Sep 5, 2019 at 20:36 comment added Nicky For ordinary category $C$, $\pi_0^{\tau}$ is the left adjoint of the inclusion $Sh(C)\to sPre(C)$, where a sheaf is regarded as discrete simplicial sheaf?
Sep 5, 2019 at 19:09 history edited Simon Henry CC BY-SA 4.0
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Sep 5, 2019 at 18:57 history edited Simon Henry CC BY-SA 4.0
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Sep 5, 2019 at 18:53 comment added Simon Henry Absolutely, I'm just not sure what language you would like, but I've added a rephrasing of the argument at the end that will be easier to translate into model theoretic or traditional homotopy theoretic language. I hope this helps. Anyway the point I'm making here is that what make thing work is not so much the description of the group completion as $\Omega B$ but rather its universal property with respect to 'group' object (in the sense monoid whose $\pi_0$ is a group) which somehow make this commutation trivial.
Sep 5, 2019 at 18:49 history edited Simon Henry CC BY-SA 4.0
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Sep 5, 2019 at 18:42 comment added Nicky Is there a way to see this without using $\infty$-category? For an ordinary category $C$, how can I see that the homotopy sheaf defined in my question is a left adjoint?
Sep 5, 2019 at 15:18 history answered Simon Henry CC BY-SA 4.0