Kan condition in simplicial homotopy theory

Question

I know Kan condition (see http://en.wikipedia.org/wiki/Kan_fibration) is something like homotopy extension condition, and I know this condition ensures homotopy defined by the naive idea to be an equivalence condition. But this condition rules out some easy-constructed simplicial sets, like those obtained from the standard simplex, and simplicial sets satisfying this condition are always quite huge.

My question is: can we avoid Kan condition by considering ALL simplicial sets and define homotopy to be an equivalence condition by ARTIFICIALLY adjoining all transitivity? Can we define well-behaved homotopy groups then for all simplicial sets and still get the equivalence between its homotopy category and that of CW complex?

Answer 1

My best guess at what you are asking is this:

Can one define homotopy groups of arbitrary simplicial sets using only simplicial homotopy?

I would say no. Indeed, consider $\partial \Delta^2$. Its geometric realisation is the circle $S^1$, so $\pi_1 (\partial \Delta^2)$ is supposed to be $\mathbb{Z}$. Recall that for a pointed Kan complex $(K, x)$, $\Omega (K, x)$ is defined by the following pullback diagram, $$\begin{array}{ccc} \Omega (K, x) & \to & [\Delta^1, K] \\ \downarrow & & \downarrow \\ \Delta^0 & \to & [\partial \Delta^1, K] \end{array}$$ where $[-, -]$ is the internal hom of simplicial sets and the bottom arrow corresponds to the constant morphism $\partial \Delta^1 \to K$ with value $x$; $\pi_1 (K, x)$ is then defined to be $\pi_0 \Omega (K, x)$. If we take this definition and apply it to $\partial \Delta^2$ we end up with a finite set – in fact, a singleton.

The problem here is that there are nowhere near enough morphisms $\Delta^1 \to \partial \Delta^2$ to represent all paths in the circle (even up to homotopy). You can try to fix this by allowing subdivisions of $\Delta^1$, but to get all the paths you will need subdivisions iterated arbitrarily many times. At that point, you may as well just apply Kan's $\mathrm{Ex}^{\infty}$ functor to $\partial \Delta^2$ and get a Kan complex.

Answer 2

You asked: 'can we avoid Kan condition by considering ALL simplicial sets and define homotopy to be an equivalence condition by ARTIFICIALLY adjoining all transitivity?' Why not do things as follows?

To turn homotopy into an equivalence relation you have to not only generate under transitivity but also to add inverses, so to do that you have to form the free groupoid ... doh! This has been done systematically by Dwyer and Kan (and also by Joyal and Tierney at about the same time) by forming the loop groupoid on the simplicial set, cf. https://ncatlab.org/nlab/show/Dwyer-Kan+loop+groupoid. Of course this ends up with a simplicially enriched groupoid (aka simplicial groupoid) and yes, of course, you can use this to calculate homotopy groups, etc. via the Moore complex construction. ... and yes of course, you can go back to simplicial sets via the loop-groupoid functors left adjoint $\overline{W}$... and low and behold that yields a Kan complex (and the fillers have the sort of properties you would like in answer to your question ... (although this is rarely shown in texts). This has an advantage in as much as it comes with a lot of fibre bundle / fibration intuitions analogous to those for the loop group in topology. These are classical, very neat and very useful!