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I was thinking about homotopy theory and classical questions such as "do homotopy groups characterize the homotopy type of a space ?", and came up with this argument :

$hTop$ (the category of topological spaces and continuous maps up to homotopy) is not concrete, but $Set^\mathbb{N}$ ($\mathbb{N}$ discrete category) is and so $Grp^\mathbb{N}$ as well, hence $(\pi_k)$ cannot be a faithful functor : there are $f, g : X\to Y$ that are not homotopic but such that for all $k$, the induced maps $f_*, g_* : \pi_k(X) \to \pi_k(Y)$ are equal.

I was wondering if in this special case, not faithful implied not conservative : could one deduce from this argument that there are non homotopy-equivalent spaces $X$ and $Y$ that have isomorphic $\pi_k$'s for all $k$ ?

Maxime Ramzi
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    Related : https://math.stackexchange.com/questions/2462958/conservativity-implies-faithfulness – Arnaud D. Apr 10 '18 at 11:36
  • Moreover that functor cannot be faithful, in fact Homotopy is not concrete: http://emis.ams.org/journals/TAC/reprints/articles/6/tr6.pdf – Ivan Di Liberti Apr 10 '18 at 12:32
  • @IvanDiLiberti : I am aware that the functor is not faithful, my question is about its conservativity – Maxime Ramzi Apr 10 '18 at 13:01
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    Beware that the last statement is not the conservativity of $\pi_$. The conservativity is : if $f:X\rightarrow Y$ is such that $f_:\pi_k(X)\rightarrow\pi_k(Y)$ is an isomorphism for all $k$, then $f$ is an isomorphism (in the homotopy category). In this post, there is examples of non homotopically equivalent CW complexes which have the isomorphic homotopy groups : https://math.stackexchange.com/questions/1954355/is-it-possible-that-isomorphic-pi-ns-not-induced-by-a-map?rq=1 – Roland Apr 10 '18 at 14:12
  • Now, just to be sure, by $hTop$, you mean the category of topological spaces up to homotopy, or $Top[w^{-1}]$ the category of topological spaces where you invert weak equivalences ? – Roland Apr 10 '18 at 14:14
  • @Roland : yes of course, it's not a reformulation of conservativity, both questions are of interest to me (but actually, proving that it's not conservative would be better to me). Thanks for the post, but I'm also trying to find out whether there can be an argument that looks like what I'm doing for "unfaithfulness". As for $hTop$ I mean the category of topological spaces and for $X,Y$, $Hom(X,Y)$ is the set of equivalence classes of continuous maps $X\to Y$ modulo homotopy – Maxime Ramzi Apr 10 '18 at 15:05
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    @Max that is not the category people mean by the phrase "homotopy is not concrete", which is the category of CW complexes and homotopy classes of maps, or equivalently the category of topological spaces with weak equivalences inverted. Your category is also non-concrete, but it's virtually never used in homotopy theory. – Kevin Carlson Apr 10 '18 at 15:35
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    @KevinCarlson : Really ? I was convinced that that was the one, thank you for the correction – Maxime Ramzi Apr 10 '18 at 15:53

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The most natural condition for a conservative functor $F$ to be faithful is that $F$'s domain should have equalizers or coequalizers which $F$ preserves. This condition isn't available in homotopy theory. There's a similar condition involving homotopy (co)equalizers, but for that you need more than just the homotopy category.

As to proving that $(\pi_k)$ is not conservative, well, the first problem is that it's not a functor, as you need to choose basepoints. If you take pointed spacee $hTop_*$, i.e. your category but with based maps and homotopies, then there are two different easy way to see $(\pi_k)$ is not conservative: there are weak equivalences which are not homotopy equivalences, and the homotopy groups see nothing about connected components away from the base point. Thus one should really consider $whTop_{*,x}$, the category of based, connected spaces with weak equivalences inverted. And in that case $(\pi_k)$ is indeed conservative, which is Whitehead's theorem.

If you wanted to stick to the unpointed case, then you could use the functor $\prod_n whTop(S^n,X)$, which is less obviously nonconservative-one trick is to consider $X$ and $Y$ as classifying spaces of groups and note that $whTop(S^1,BG)$ is the set of conjugacy classes of elements of $G$. But in any case, one has to do some kind of a direct proof, unrelated to faithfulness. And there can be no general abstract argument that $whTop$ (or $hTop,hTop_*$, etc) admits no conservative functor into $Sets$, as with the result that homotopy is not concrete, as every locally small category admits such a functor.

Kevin Carlson
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  • Yes sorry, I meant $hTop_*$, so with basepoints (otherwise $\pi_k$ is not even defined, unless maybe as a groupoid); and similarly indeed, I should have considered only path connected topological spaces. But then it is conservative ? Wow I did not expect that. I just looked up Whitehead's theorem, so could you perhaps expand on the connection between the category I described in the comments -modulo the modifications mentioned here- and the one with CW-complexes (that Whitehead's theorem is about) ? Or the one with weak equivalences inverted ? Thank you – Maxime Ramzi Apr 10 '18 at 15:58
  • @Max Sure. Every topological space is weakly equivalent to a CW-complex, for instance, the geometric realization of its singular simplicial set. So the category of spaces, localized at weak equivalences, is equivalent to the category of CW complexes localized in the same way; in both cases the functor $(\pi_k)$ is conservative by definition. What's better is Whitehead's theorem says that the localizations of CW complexes at weak equivalences and at homotopy equivalences are equivalent. So $(\pi_k)$ is also conservative on CW complexes with homotopy classes of maps. – Kevin Carlson Apr 10 '18 at 20:09
  • On the other hand, such a conservative functor do exists, and is close to be a generalized collection of $\pi_n$. – Ivan Di Liberti Apr 10 '18 at 22:45
  • @IvanDiLiberti really? The functor of Freyd I would think we're both thinking of doesn't strike me as anything like a collection of $\pi_n$s. – Kevin Carlson Apr 10 '18 at 23:09
  • It is in some sense a collection of well chosen sub objects. – Ivan Di Liberti Apr 11 '18 at 06:05
  • Yes, it's kind of a quotient of a subfunctor of the disjoint union of all the representables at once, which seems too far from a disjoint union of a small set of representables. But I guess this is a matter of opinion. – Kevin Carlson Apr 11 '18 at 15:29