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Mar 22, 2023 at 0:40 vote accept user494312
Mar 21, 2023 at 4:38 comment added Reid Barton What I meant is that, for fixed X, your equation holds for all Y if and only if the counit |Sing X| -> X is a homeomorphism. And both |Sing X| and (by assumption) X are "nice" spaces, so if (as usual) they are different, we should be able to see the difference using maps into another "nice" space Y.
Mar 20, 2023 at 15:05 comment added user494312 Thank you. Though, is there a positive result saying that any counterexample is of this kind ? Say, comes from a natural transformation $|-|\implies|-|$ or something like that (actually, I do not think that your conterexample does arise in this way, so what I suggest is not enough.). In other words, to make formal your remark "Your question is implicitly about the counit $|SingX| \to X$."
Mar 20, 2023 at 14:46 comment added LSpice The failure (though not the counterexample) is also mentioned in a comment of @ChrisSchommer-Pries.
Mar 20, 2023 at 7:07 history answered Reid Barton CC BY-SA 4.0