Closure of the homotopy relation for a simplicial set - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:44:56Z http://mathoverflow.net/feeds/question/29633 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29633/closure-of-the-homotopy-relation-for-a-simplicial-set Closure of the homotopy relation for a simplicial set roger123 2010-06-26T18:58:14Z 2010-06-27T03:26:42Z <p>Define a reflexive relation on the set of zero-simplices of a simplicial set $A$ by saying that $x\sim y$ iff there is a one-simplex $h$ with $0$-face $y$ and $1$-face $x$. This is not an equivalence relation in general but it is so if $A$ is Kan. Define $\pi_0(A)=A_0/\sim$ in this case.</p> <p>Let $x:\Delta^0\to A$ be a zero-simplex, $n\in\mathbb{N}$ and consider the pullback <code>$\begin{array}{rcl} A(n,x) &amp;\to&amp; Map(\Delta^n,A)\\ \downarrow &amp;&amp;\downarrow\\ Map(\partial\Delta^n,\Delta^0)&amp;\to&amp;Map(\partial\Delta^n,A) \end{array}$</code> induced by the obvious maps and set $\pi_n(A,x)=\pi_0(A(n.x))$ as the simplicial homotopy groups. This works since $A(n,x)$ is Kan if $A$ is so.</p> <p>An important theorem states that $\pi_n(A,x)=\pi_n(|A|,|x|)$ the bars denoting the realization functor adjoint to the singular functor $S$. The homotopy category of the usual model structure on simplicial sets is given by inverting the "weak equivalences", i.e. the maps $f:A\to B$ such that $\pi_n(S(f),x)$ is an isomorphism for all $n$ and all basepoints. One has to apply $S$ here to make things Kan.</p> <p>Does one get the same homotopy category if one lets $\pi_0$ be the equivalence classes of the equivalence relation <strong>generated</strong> by $\sim$? One can define all necessary concepts exactly as above without demanding $A$ to be Kan. Does one get the same homotopy category then?</p> http://mathoverflow.net/questions/29633/closure-of-the-homotopy-relation-for-a-simplicial-set/29655#29655 Answer by Tom Goodwillie for Closure of the homotopy relation for a simplicial set Tom Goodwillie 2010-06-27T01:27:07Z 2010-06-27T03:26:42Z <p>Of course if $\pi_0$ is defined by the equivalence relation generated by ~ on $0$-simplices then it is the usual thing: topological $\pi_0$ of the realization, or simplicial $\pi_0$ of a fibrant replacement.</p> <p>You are saying: What if we define a new $\pi_n(A,x)$ as $\pi_0(A(n,x))$? Well, obviously it maps to the usual $\pi_n(A,x)=\pi_0(S(A),n,x)$), and clearly this map is rarely an isomorphism if $A$ is not fibrant. But I don't even see a comparison map between the resulting homotopy category and the usual one, in either direction. Clearly a map of simplicial sets will sometimes induce an isomorphism of usual homotopy groups while inducing a non-isomorphism of yours. But (this is my point) it can also go the other way. For example, you can make lots of examples of simplicial sets $A$ such that the inclusion $V\to A$ of the $0$-skeleton of $A$ induces an isomorphism $V(n,x)\to A(n,x)$.</p>