Given $f:X \to Y$ a continuous map between two spaces (unpointed CW-complexes) such that $f$ induces an isomorphism in homology with integer coefficient, and $f$ induces an isomorphism on homology of the free loop spaces: $H_*(X^{S^1}) \to H_*(Y^{S^1})$ (also with integer coefficient). Does $f$ has to be a weak homotopy equivalence ?

If I further assume that $f$ induces an equivalence $H_*(X^{K}) \to H_*(Y^{K})$ for each finite CW-complex $K$, is $f$ a weak homotopy equivalence ?

Note that working in the unpointed settings makes a big difference here: It was showed by Arlin and Christensen that, contrary to the case of pointed connected space, for any small set of spaces $E$, there are maps that induces an isomorphism $\pi_0(X^K) \to \pi_0(Y^K)$ for all $K \in E$ without being weak equivalences. It is unclear to me if adding higher homology groups makes a big difference or not.

Edit: The case of groupoids $X=BG$ and $Y=BH$ is already an interesting example. $X^{S^1}$ is the (classyfing space of) the groupoid corresponding to the action of $G$ on itself by conjugation. Hence given $f:G \to H$ a morphism of group. Saying that it is a bijection on the $H_0$ of the loop space gives that $f$ induces a bijection between the set of conjugation classes of $G$ and $H$, and the fact that it is a bijection on the $H_1$ of the loop spaces gives that for all $g \in G$, $f$ induces an isomorphisms between the ablianization of the centralizer of $g$ and the abelianization of the centralizer of $h$.

That does not seem to be quite enough to conclude that $f$ is an isomorphisms, but this is already pretty restrictive. I havn't been able to formulate the fact that $f$ induce bijections on the higher $H_i$ in simple terms.

Now, if we start looking a $H_0(X^K)$ and $H_0(Y^K)$ for $K$ a finite complex, it seems that in this case all the interesting information is in the case where $K$ is a wedge of circle. Being a bijection on $H_0(X^K)$ and $H_0(Y^K)$ means that:

1) $f$ induces a bijection between $G^n/G$ and $H^n/H$ for all $n \in \mathbb{N}$, where $G$ and $H$ acts on $G^n$ and $H^n$ by the diagonal conjugation action.

2) For every finite family of elements $g_1,\dots,g_n$ in $G$, $f$ induce a bijection between the abelianization of the centralizer of $\{g_1,\dots,g_n\}$ and the abelianization of the centralizer of $\{f(g_1),\dots,f(g_n) \}$.

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    $\begingroup$ For your first question, what if $X=BG$ and $Y=pt$, with $G$ an acyclic group ? $\endgroup$
    – GSM
    Apr 26, 2020 at 20:00
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    $\begingroup$ That does not seems enough to me: $BG^{S^1}$ is the homotopy quotient of the action of $G$ on itself by conjugation. So just its $H_0$ and $H_1$ already remember the set of conjugation class of $G$ and the abelianization of the centralizer of each elements. $\endgroup$ Apr 26, 2020 at 22:46

1 Answer 1


I believe these questions were studied and answered in the this 1984 paper:

HIROSHIMA MATH. J. 14 (1984), 359-369 On the set of free homotopy classes and Brown's construction

Takao MATUMOTO, Norihiko MINAMI and Masahiro SUGAWARA

They also have counterexamples that seem to be the same as in the recent preprint of Arlin and Christensen that you mention.

Take a look and see if they do what you want.

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    $\begingroup$ Just to clarify, the example in this paper is well known, and is what we generalized beyond finite complexes in our paper. $\endgroup$ May 1, 2020 at 20:33
  • $\begingroup$ I might be missing something, but I don't understand how this paper answer the question ? They give a counter-exemple to the case where we look at free homotopy class [K, _ ] for K a finite CW-complex (which was already mentioned in the question), but I don't think it is a counter-example to the case where we also look at homology group, and it proves that it works when considering [K,X] when K is an arbitrary CW-complexes, whose special case relevant for the question is clear by Yoneda lemma ? $\endgroup$ May 1, 2020 at 21:02
  • $\begingroup$ But their theorem 1 is definitely relevant: it shows that it is enough to show that $f$ is surjective on $\pi$ to conclude that it is an equivalence. $\endgroup$ May 1, 2020 at 21:03

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