# recognising weak equivalences of simplicial sets

$\require{AMScd}$ I am interested in detecting using a lifting property when a map $f:X\to Y$ in $sSet$ (with the standard Kan model structure) is a weak equivalence. In the paper Weak Equivalences and Simplicial Presheaves (http://www.math.uiuc.edu/K-theory/0564/wesp.pdf) Daniel Dugger and Daniel Isaksen give the following criterion (due to Reedy):

A map $f:X\to Y$ between fibrant simplicial sets is a weak equivalence if and only if it has the relative homotopy lifting property (RHLP, see below) with respect to all generating cofibrations $\partial\Delta^n\hookrightarrow \Delta^n.$

They also mention that a similar result would hold for non-fibrant objects $X,Y$ if one allows to subdivide $\partial \Delta^n$ and $\Delta^n.$ What exactly would this lifting criterion for general maps $f:X\to Y$ be?

Here is the definition of the RHLP: A square $$\begin{CD} A @>>> X\\@VVV@VVV\\B@>>>Y\end{CD}$$ has a relative homotopy lifting if there exist a lift $B\to X$ making the upper triangle commute on the nose and a simplicial homotopy relative to $K$ from $B\to X\to Y$ to the given map $B\to Y.$ The map $X\to Y$ has the RHLP with respect to $A\to B$ if every such square has a relative homotopy lifting.


$$\begin{CD} X @>{\sim}>> \Ex^\infty X \\ @V{f}VV @VV{\Ex^\infty f}V \\ Y @>>{\sim}> \Ex^\infty Y \textrm{.} \\ \end{CD}$$

The map $f$ is a weak equivalence if and only if $\Ex^\infty f$ is if and only if there is a relative lift in every diagram of the form

$$\begin{CD} \partial\Delta[n] @>>> \Ex^\infty X \\ @VVV @VV{\Ex^\infty f}V \\ \Delta[n] @>>> \Ex^\infty Y \textrm{.} \\ \end{CD}$$

Since such a lifting problem and its solution involve only finitely many simplices of $\Ex^\infty X$ and $\Ex^\infty Y$, there is a $j$ such that the lifting problem actually lives in

$$\begin{CD} \partial\Delta[n] @>>> \Ex^j X \\ @VVV @VV{\Ex^j f}V \\ \Delta[n] @>>> \Ex^j Y \\ \end{CD}$$

and a $k$ such that its solution lives in

$$\begin{CD} \partial\Delta[n] @>>> \Ex^j X @>>> \Ex^{j+k} X\\ @VVV @VV{\Ex^j f}V @VV{\Ex^{j+k} f}V \\ \Delta[n] @>>> \Ex^j Y @>>> \Ex^{j+k} Y \textrm{.} \\ \end{CD}$$

By adjointness, this means that $f$ is a weak equivalence if and only if for every $j$ and every lifting problem

$$\begin{CD} \Sd^j\partial\Delta[n] @>>> X \\ @VVV @VV{f}V \\ \Sd^j\Delta[n] @>>> Y \\ \end{CD}$$

there is a $k$ and a "relative lift" in

$$\begin{CD} \Sd^{j+k}\partial\Delta[n] @>>> \Sd^j\partial\Delta[n] @>>> X \\ @VVV @VVV @VV{f}V \\ \Sd^{j+k}\Delta[n] @>>> \Sd^j\Delta[n] @>>> Y \textrm{.} \\ \end{CD}$$

However, you have to be careful here since a homotopy $\Delta[n] \times \Delta[1] \to \Ex^{j+k} Y$ translates to a map $\Sd^{j+k}(\Delta[n] \times \Delta[1]) \to Y$ which is different than $\Sd^{j+k}\Delta[n] \times \Delta[1] \to Y$ or $\Sd^{j+k}\Delta[n] \times \Sd^{j+k}\Delta[1] \to Y$ so "relative lift" has a slightly different meaning now.

• But the homotopy need not factor through the same Ex^k, it might land in some Ex^l where l is much bigger than k. So in the lifting condition one must allow k to increase first. – Dmitri Pavlov Jan 29 '15 at 15:37
• A relative lift involves a choice of a homotopy, so my $k$ is already your $l$. – Karol Szumiło Jan 29 '15 at 15:44
• Thank you, but your answer seems to assume that every map $\partial\Delta^n\to {Ex}^\infty X$ factorizes as $\partial\Delta^n\to X\to {Ex}^\infty X$ and that this factorization is functorial in some sense. Why is this always true? – COhrt Jan 29 '15 at 16:43
• That's not true and I'm not using that, I'm only using that there is a factorization $\partial\Delta[n] \to \mathrm{Ex}^k X \to \mathrm{Ex}^\infty X$ for some $k$. – Karol Szumiło Jan 29 '15 at 16:58
• Then I don't understand how in your first step it is enough to only show that $Ex^\infty f$ admits a lift in the squares factorizing through $X$ and $Y$ to show that it is a weak equivalences. Wouldn't we have to show that $\Ex^\infty f$ admits a lift in all squares involving $\partial \Delta^n\to \Delta^n$? – COhrt Jan 29 '15 at 17:16