The question is related to the question: detecting weak equivalences in a simplicial model category

Suppose that we have a simplicial model category $M$ and denote by $M^{f}$ the full simplicial subcategory of fibrant objects. Suppose that $R$ is a subcategory of $M^{f}$ such that for any object $m\in M^{f}$ there exists an object $r\in R$ such that $r$ is (zigzag) equivalent to $m$ i.e. $r$ and $m$ are isomorphic in $Ho(M)$ the homotopy category of $M$. Let $w: a\rightarrow b$ be a morphism in $M$ such that for any object $r\in R$ the induced map of simplicial sets $w^{\ast}:Map_{M}(b,r)\rightarrow Map_{M}(a,r)$ is a weak homotopy equivalence of simplicial sets. Can we conclude that $w$ is a weak equivalence in the model category $M$ ?

If $R=M^{f}$ this is true and is proved in Hirschhorn's book.

EDIT : The question is very general, and it should have a formal answer in case it is true. After trying all suggestions (in the comments and deleted answer) and reading Hirshorn's book I had the impression that maybe the answer to my question is no, and there should be a counterexample. In the Hirshorn's book the fact we test for all fibrant objects seems to be essential.

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    $\begingroup$ Unless I've misunderstood, essentially the same argument as in Hirschhorn's book should give an affirmative answer to your question. (First do a cofibrant-fibrant replacement of a and b, then do a fibrant replacement of w; apply the argument to this replacement of w.) $\endgroup$
    – skd
    Sep 4, 2018 at 21:29
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    $\begingroup$ @skd I'm a beginner in the field, I tried your suggestion, I was unable to provide a correct proof. I feel I'm missing some thing... Any help with more detail is welcome. $\endgroup$
    – Let
    Sep 5, 2018 at 7:27
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    $\begingroup$ I agree with OP. The question is not that simple and the answer might as well be no. It seems that @skd's suggestion doesn't work because we need to take cofibrant fibrant replacements of a and b which belong to $R$, but I think we can't choose such a replacement in general. $\endgroup$ Sep 6, 2018 at 16:36
  • $\begingroup$ Probably this can be proven if every object is cofibrant. So when looking for counterexamples, it's probably best to look at simplicial model categories where not every object is cofibrant. $\endgroup$
    – Tim Campion
    Sep 7, 2018 at 2:55

1 Answer 1


I think the following gives a counterexample:

Take the category of morphisms in $\operatorname{SSet}$, i.e., the category $\operatorname{Fun}(C,\operatorname{SSet}),$ where $C$ is the category with two objects and one non-identity morphism. Equip this category with the projective model structure. $R$ will be the category of morphisms $f:X\rightarrow Y$ where $X,Y$ are Kan complexes and $f$ is a $\operatorname{SSet}$-cofibration (i.e., monomorphism.)

Let $a$ correspond to the morphism $\operatorname{Sing}(S^1)\rightarrow\Delta^0$ and let $b$ correspond to the morphism $\Delta^0\rightarrow\Delta^0.$

If I haven't made a mistake, with these choices, $\operatorname{Map}(b,r)\rightarrow\operatorname{Map}(a,r)$ is even an isomorphism for all $r\in R$. Every fibrant object in $\operatorname{Fun}(C,\operatorname{SSet})$ is equivalent to an object of $R$ (factor your morphism into a cofibration+trivial fibration and take the trivial fibration part.)


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