$\require{AMScd}$ 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.


1 Answer 1


$\newcommand{\Ex}{\mathrm{Ex}}\newcommand{\Sd}{\mathrm{Sd}}$Consider the square

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

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.

  • $\begingroup$ 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. $\endgroup$ Jan 29, 2015 at 15:37
  • 1
    $\begingroup$ A relative lift involves a choice of a homotopy, so my $k$ is already your $l$. $\endgroup$ Jan 29, 2015 at 15:44
  • $\begingroup$ 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? $\endgroup$
    – COhrt
    Jan 29, 2015 at 16:43
  • $\begingroup$ 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 $k$. $\endgroup$ Jan 29, 2015 at 16:58
  • $\begingroup$ 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 $\partial \Delta^n\to \Delta^n$? $\endgroup$
    – COhrt
    Jan 29, 2015 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.