Consider the model structure on simplicial sets where the cofibrations are given by monomorphisms and the weak equivalences are given by $n$-equivalences (that is, maps $f \colon X \to Y$ that induce a bijection $f_{*} \colon \pi_{0}X \to \pi_{0}Y$ and isomorphisms $f_{*} \colon \pi_{k}(X,x) \to \pi_{k}(Y,fx)$ for all $0 < k \leq n$ and all $x \in X$).

This can be constructed variously as the left Bousfield localization of the standard Kan-Quillen model structure on simplicial sets, or as a Cisinski model structure where the set of generating anodyne extensions is given by: $$J_{n} = \left\{\Lambda^{k}_{i} \hookrightarrow \Delta^{k} \mid k > 0, i = 0, \ldots, k \right\} \cup \left\{\partial\Delta^{k} \hookrightarrow \Delta^{k} \mid k \geq n+2 \right\}$$

We know, for instance, that the fibrant objects are $n$-truncated Kan complexes. We also know that a map with fibrant codomain is a fibration if and only if it has the right lifting property against every element of $J_{n}$.

My question is this: do we know whether or not an arbitrary map with right lifting property against every element of $J_{n}$ is a fibration in this model structure? If not, is there a simple counterexample that proves it to be false?

  • $\begingroup$ It may be helpful to observe that a map of simplicial sets has the right lifting property with respect to the set $J_n$ if and only if it is a Kan fibration whose fibres are $n$-types. $\endgroup$ Commented Apr 23 at 6:34

1 Answer 1


No, this is not true. A counterexample is provided in Hirschhorn's book, Example 2.1.6 on page 36. See also the text on page 71: "Unfortunately, Example 2.1.6 shows that not all S-local trivial cofibrations need be $J_n$-cofibrations, and so there may be $J_n$-injectives (i.e., maps having the RLP) that are not S-local fibrations." I've modified Hirschhorn's notation to match the notation in the question.

This failure is also studied in depth in Barwick's paper "On left and right model categories and left and right Bousfield localizations." The point about maps with fibrant codomain behaving well is related to semi-model categories. Not every semi-model category is a full model category. Hirschhorn's example (credited to Bousfield) shows that you can't drop this condition in general.

Unfortunately, I'm on my way off MathOverflow, so if you want to talk more, just email me.

  • $\begingroup$ Note that Hirschhorn's example concerns topological spaces, and so does not directly answer the question at hand, which is about simplicial sets. $\endgroup$ Commented Apr 23 at 5:58
  • $\begingroup$ Hi David, thanks for your answer! I believe there is a crucial "not" missing from the Hirschhorn quote. $\endgroup$ Commented Apr 23 at 10:01
  • $\begingroup$ @UditMavinkurve Thanks, I fixed that just now. $\endgroup$ Commented Apr 23 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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