Homotopic quotients of simplicial sets as infinity-groupoids - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:00:13Z http://mathoverflow.net/feeds/question/39802 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39802/homotopic-quotients-of-simplicial-sets-as-infinity-groupoids Homotopic quotients of simplicial sets as infinity-groupoids David Carchedi 2010-09-23T23:22:07Z 2011-09-17T17:03:13Z <p>Suppose $f:X \to Y$ is a function of sets. Then we can take the quotient $X/\text{~}$ by identifying $x \text{~} y$ if and only if $f(x)=f(y)$. Now suppose instead that $f:X \to Y$ is a map of simplicial sets. I want to emulate this homotopically, by adding a 1-simplex between $x$ and $y$ if there is a 1-simplex from $f(x)$ to $f(y)$, (and similarly on higher simplices). This is probably most clear if you think of $X$ and $Y$ as infinity groupoids (as indeed I have in mind). I want a way of "making guys equivalent if they're equivalent after applying f". So, if $X$ and $Y$ are Kan, adding a 1-simplex between two 0-simplices makes them "weakly isomorphic" (which is the correct thing to do, not just glue them together outright). Is there a standard construction for maps of (maybe Kan?) simplicial sets that does this?</p> http://mathoverflow.net/questions/39802/homotopic-quotients-of-simplicial-sets-as-infinity-groupoids/39811#39811 Answer by David Roberts for Homotopic quotients of simplicial sets as infinity-groupoids David Roberts 2010-09-24T00:57:26Z 2010-09-24T00:57:26Z <p>You are looking at the coequaliser of the kernel pair, so my guess would be to take the homotopy pullback of $f$ along itself, then look at the nerve of the groupoid $X\times_Y X \rightrightarrows X$ in $sSet$ this gives rise to, then form the diagonal (=hocolim) of this bisimplicial set. I guess this comes with a map to $Y$, but I haven't checked.</p> http://mathoverflow.net/questions/39802/homotopic-quotients-of-simplicial-sets-as-infinity-groupoids/39829#39829 Answer by Tim Porter for Homotopic quotients of simplicial sets as infinity-groupoids Tim Porter 2010-09-24T07:16:26Z 2010-09-26T06:51:57Z <p>This answer is perhaps a gloss on David's one. It is often useful to replace taking a quotient by forming the equivalence relation as a groupoid. Thus the initial situation you describe has the classical equivalence-relation-from-a-function form. This will work in any category with pullbacks as it is the pullback of f along itself. In a homotopy situation, such as you need, the analogue will be the homotopy pullback of $f$ along itself.</p> <p>This does not form the quotient as such, but is, I maintain, better (especially in the presence of differential structures for instance).It corresponds to the idea that was sketched in the question, but is natural functorial and so less hassle(&lt;- technical categorical term meaning 'less hassle'!). It is also going to give results that do not depend on the homotopy class of $f$ and that is often important especially if you are thinking of the simplicial sets as being weak infinity groupoids or similar. I believe there are extensions to quasicomplexes but do not have sources with me to check at the moment or to give chapter and verse. </p> <p>This construction not only says two simplices in $Y$ are to be thought of as being the same but records WHY, and that is important.</p> <p>(Edit: Thanks Tom. I should have said 'It is also going to give results that only depend on the homotopy class of $f$ ..')</p> http://mathoverflow.net/questions/39802/homotopic-quotients-of-simplicial-sets-as-infinity-groupoids/39849#39849 Answer by Urs Schreiber for Homotopic quotients of simplicial sets as infinity-groupoids Urs Schreiber 2010-09-24T11:56:57Z 2010-09-25T10:06:28Z <p>I'd think the construction in question is the <a href="http://ncatlab.org/nlab/show/coimage#InInfCat" rel="nofollow">homotopy coimage</a> of $f$ (it's unfortunately called "coimage" even though it behaves like the image).</p> <p>First one forms the homotopy Cech nerve</p> <p>$$<br> C(f) = \left( \cdots X \times_Y X \times_Y X \stackrel{\to}{\stackrel{\to}{\to}}X \times_Y X \stackrel{\to}{\to} X \right)$$</p> <p>This is the internal groupoid object that encodes the $\infty$-equivalence relation on the elements of $X$ "is equivalent in $Y$"</p> <p>Forming its homotopy colimit</p> <p>$$coim(f) := \lim_{\to} C(f)$$</p> <p>produces the homotopy quotient of $X$ by this equivlence relation.</p> <p>As an example, take $\mathbf{B}G$ the one-object groupoid of a group $G$ and $* \to \mathbf{B}G$ the point inclusion. One wants to see that the homotopy image of the point inclusion $* \to \mathbf{B}G$ is not just the point, but is $*//G$, i.e. $\mathbf{B}G$ itself, because there is $G$ worth of ways for the point to be equivalent to itself after inclusion into $\mathbf{B}G$.</p> <p>So one computes the homotopy Cech nerve and finds the familiar</p> <p>$$C(* \to \mathbf{B}G) = \left(<br> \cdots G \times G \times G \stackrel{\to}{\stackrel{\to}{\to}} G\times G \stackrel{\to}{\to} G \right)$$</p> <p>but now regarded as a simplicial object in $\infty Grpd$. This is the diagram that encodes the action of $G$ on the point. Its homotopy colimit is indeed again $\mathbf{B}G$, so that we find</p> <p>$$coim(* \to \mathbf{B}G) = \mathbf{B}G$$</p> <p>That this comes out this way is an example of Giraud's axioms at work, since $\infty Grpd$ happens to be an $\infty$-topos: this implies that <em>every</em> $\infty$-groupoid object (simplicial diagram as above) in $\infty Grpd$ is <em>effective</em> .</p> http://mathoverflow.net/questions/39802/homotopic-quotients-of-simplicial-sets-as-infinity-groupoids/75685#75685 Answer by Ronnie Brown for Homotopic quotients of simplicial sets as infinity-groupoids Ronnie Brown 2011-09-17T17:03:13Z 2011-09-17T17:03:13Z <p>You might like to look at the 1978 thesis of Nick Ashley on "Simplicial $T$-complexes and crossed complexes: a nonabelian version of a theorem of Dold and Kan." available from Esquisses Math. 1978 at <a href="http://ehres.pagesperso-orange.fr/Cahiers/Ctgdc.htm" rel="nofollow">http://ehres.pagesperso-orange.fr/Cahiers/Ctgdc.htm</a></p> <p>He considers a filtered Kan complex $K_*$ and a natural homotopy relation to give a Kan fibration $p: K_* \to \rho(K_* )$, where $\rho(K_*)$ is a simplicial $T$-complex, i.e. a strong form of Kan complex with unique "thin" fillers. Modifications of this should give you strict consructions of the kind you want. </p>