Let $\mathcal{H}$ denote the homotopy category associated with the KanQuillen model structure on $\mathbf{sSet}$. Suppose we have a map $f\colon X \to Y$ between Kan complexes, such that for every finite simplicial set $K$ we have an isomorphism of the form: $$\mathcal{H}(K,X) \cong \mathcal{H}(K,Y)$$ induced by postcomposition with $[f]$. Is it true that $f$ is then a weak equivalence?

2$\begingroup$ I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version. $\endgroup$ – user123627 Oct 12 '18 at 5:58

1$\begingroup$ You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $\pi_n$ by the action of $\pi_1$. So it at least works in cases where the $\pi_1$ action is trivial $\endgroup$ – Simon Henry Oct 12 '18 at 6:20

2$\begingroup$ Lawson’s counterexample of the shift map $s$ on the classifying space $X$of the restricted infinite symmetric group for spaces also applies to simplicial sets, since the equivalence between the two categories sends finite simplicial sets to CW complexes and vice versa, up to weak equivalence. It is indeed true, as Simon points out, that any finite family of circles in $X$ is sent by $s$ to a family of circles representing simultaneously conjugate classes in the fundamental group. But this isn’t enoughthere are infinite families not simultaneously conjugate to their shifts. $\endgroup$ – Kevin Arlin Oct 12 '18 at 7:41

1$\begingroup$ Cont’d : In fact there are no small families of objects in the homotopy category for which the analogous claim is true. Similar counterexamples go through for classifying spaces of arbitrarily big restricted symmetric groups. This is a way in which the unpointed homotopy category is unavoidably terrible compared to the pointed or stable categories. $\endgroup$ – Kevin Arlin Oct 12 '18 at 7:47

3$\begingroup$ @SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups. $\endgroup$ – Kevin Arlin Oct 12 '18 at 8:02
The answer is no. Otherwise, it would follow from Brown's representability theorem (and here I mean very specifically Theorem 2.8 from Brown's 1965 paper Abstract Homotopy Theory) that every "halfexact" functor on the homotopy category of unbased simplicial sets is representable. That is however false by my answer here. See also this post and the discussion below for some context.
Nice question!
I would suspect that the existence of phantom maps (see the nLab entry on these for starters) would suggest that the answer is a lot more complex than one would suspect. I quote
''a continuous map $f$ from a CWcomplex $X$ to a topological space $Y$ is a phantom map if it is not homotopic to a constant but for every finite CWsubcomplex $Z\subset X$ the restriction $f_Z:Z\to Y$ is homotopic to a constant."
The conditions you impose would imply that $X$ and $Y$ have the same $n$type for all $n$. From an example of Adams in 1957, and in reply to an original question of J. H. C. Whitehead (both looking at CWcomplexes rather than simplicial sets), we get that $X$ and $Y$ are not necessarily homotopy equivalent. (Do a search on `Spaces of the same ntype for all n' to get some papers relevant to this.) The obstruction is in a $lim^{(1)}$ group associated to the automorphisms of the spaces.
This is not quite an answer to that question since you are specifying that $f$ is given to start with, but is clearly related to your and Harry's doubts given that you are asking for unpointed mapping spaces.
A related problem occurs in Proper Homotopy Theory and there perhaps you can find a nice conceptual example as there are geometric interpretations of the $lim^{(1)}$ groups. The intuition is that the homotopies needed to give the isomorphisms $$\mathcal{H}(K,X) \cong \mathcal{H}(K,Y)$$ might not be made homotopically coherent enough to build a homotopy inverse.
Clearly in your case you need $X$ and $Y$ to have nontrivial homotopy groups in all dimensions.
I hope this helps.

$\begingroup$ Is it true if you ask instead for the induced map $X^K\to Y^K$ on simplicial mapping spaces to be a weak equivalence for all finite simplicial sets $K$? $\endgroup$ – Harry Gindi Oct 12 '18 at 6:15

4$\begingroup$ Take $K=\Delta[0]$ then aren't you putting the conclusion in as a condition! or is that not what you are intending. $\endgroup$ – Tim Porter Oct 12 '18 at 6:35

$\begingroup$ Thanks for your answer Tim. I know that having the same homotopy groups is not enough, but in this case (as you point out) a map inducing the iso is given, so I think I am not sure I understood your message. $\endgroup$ – Edoardo Lanari Oct 12 '18 at 7:02


$\begingroup$ It’s not actually the case that $X$ and $Y$ need to be homotopically infinite dimensional heresee the counterexample among classifying spaces of discrete groups linked in comments. $\endgroup$ – Kevin Arlin Oct 12 '18 at 7:52