Dtyjlui

I am trying to understand a specific construction of Postnikov towers for simplicial sets, as explained for instance here (under "absolute Postnikov tower")

So you start with a simplicial set $X$ (I don't know if it works for all simplicial sets, I don't mind assuming it's a Kan complex if it's necessary; or even if it makes the argument substantially easier) and define, for all $ngeq 0$ the following relation on simplices of $X$: if $alpha,beta in X_q$, seen as $alpha,beta :Delta^q to X$ , then $alphasim_n beta$ if and only if $sk_n(alpha) = sk_n(beta) : sk_nDelta^q to X$.

It is quite clear that this relation is compatible with boundaries and degeneracies, because if you look at $alpha in X_q$ as $alpha: Delta^qto X$, then $d_i(alpha) = alphacirc delta_i$, and $sk_n$ is a functor so everything works well here; and we may define $X(n) = X/sim_n$ as a simplicial set; and we have canonical projection maps $Xto X(n), X(n+1)to X(n)$ for all $ngeq 0$.

It's easy to see that the $X(n)$ satisfy the homotopy groups part of the requirements for a Postnikov tower; but I'm struggling to see why $X(n+1)to X(n)$ should be a Kan fibration. My question is :

Why is it a Kan fibration ?

Here's what I worked out : Let $p:X(n+1)to X(n)$ the wannabe-fibration; and start with $(m-1)$-simplices $x_0,...,widehat{x_k},...,x_min X(n+1)_{m-1}$ with compatible boundaries, and $yin X(n)_m$ such that $d_i(y) = p(x_i)$ for $ineq k$.

Then as $p$ is surjective, find $tilde{x} in X(n+1)_m$ such that for all $ineq k$, $d_i(tilde{x}) sim_n x_i$.

If $m-1 leq n$, then this implies $d_i(tilde{x}) = x_i$ for $ineq k$, so $tilde{x}$ is an appropriate filling. But we must now deal with the case $m-1> n$, i.e. $m-1geq n+1$.

Now this $n+1$ sort of makes me happy because in $X(n+1)$ we're only dealing with things up to their $(n+1)$-skeleton; but I don't really know what to do with it...

If we try to take $Delta^{n+1}overset{alpha}{to} Delta^{m-1} overset{x_i}{underset{d_i(tilde{x})}{rightrightarrows}} X(n+1)$ and show that they are equal, then if $alpha$ in its normal form has some codegeneracies, then we are done because then we pass to the $n$-skeleton, but if $alpha$ is only made up of cofaces, I don't know what to do.. I don't even know whether the same $tilde{x}$ will work, or if I somehow have to cook up another $z$ (and if I have to, this makes me think that perhaps I actually need $X$ to be a Kan complex, because otherwise I don't know how to "make simplices appear"); so that's why I need help.

EDIT : I noticed that I didn't mention that I knew how to prove that $Xto X(n)$ was a Kan fibration if $X$ is a Kan complex. Maybe that can help ?

EDIT 2 : Update, thanks to the first edit, I know it suffices to prove that if $qcirc p$ and $p$ are surjective fibrations, then $q$ is a fibration. For that I reduced it to the following statement (which I saw online and so think is true) : if $p: Xto Y$ is a surjective fibration, then any map $Lambda^n_k to Y$ lifts to $Lambda^n_k to X$; I'll try to prove this; but unless you have a better proof for the old question, this should be the new question : Why is this lifting property true ?