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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.

Let $x:\Delta^0\to A$ be a zero-simplex, $n\in\mathbb{N}$ and consider the pullback \[ \begin{array}{rcl} A(n,x) &\to& Map(\Delta^n,A)\\ \downarrow &&\downarrow\\ Map(\partial\Delta^n,\Delta^0)&\to&Map(\partial\Delta^n,A) \end{array} \] 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.

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.

Does one get the same homotopy category if one lets $\pi_0$ be the equivalence classes of the equivalence relation generated 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?

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Yes. More than that: I want to say that when you ask if the two categories are the same, you must really be asking if they are isomorphic (or perhaps equivalent as categories?). But since they have the same objects, it would be pretty weird if this were true other than by an isomorphism which is the identity on objects. If there were an obvious functor between them which was the identity on objects, then this would be a more reasonable question would be 'is it an isomorphism?'; but I pointed out that there is no functor in either direction making a commutative triangle. –  Tom Goodwillie Jun 27 '10 at 20:41
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I mean, making a commutative triangle with the given functors {usual weak homotopy cat} <-- simplicial sets --> your homotopy cat –  Tom Goodwillie Jun 27 '10 at 20:42

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up vote 5 down vote accepted

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.

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 $V(n,x)\to A(n,x)$.

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Thank you for your answer. So you are saying (hopefully I have understood you right) that it is most likely false? –  roger123 Jun 27 '10 at 18:21

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