I'm confused by the following question: $f:X\to Y$ is a weak homotopy equivalence, that is $f_*:\pi_*(X)\to \pi_*(Y)$ is an isomorphism for any dimensional homotopy groups. However, for the stable homotopy groups, is the homomorphism $f_*:\pi_*^s(X)\to \pi_*^s(Y)$ still an isomorphism?

Any comments are welcome! Many Thanks!

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    $\begingroup$ "Can" doesn't seem like the word you want here. Maybe "must"? Anyway, suspension is a homotopy colimit, so if there's any justice in the world then it should preserve weak equivalences. $\endgroup$ – Qiaochu Yuan Nov 15 '13 at 9:34
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    $\begingroup$ A good example to consider is where $Y=\prod_{k=1}^\infty S^k$ and $X=\bigcup_n\prod_{k=1}^nS^k\subset Y$ and $f$ is the inclusion. Then $f$ is a weak equivalence but not a homotopy equivalence, and $X$ has a degenerate basepoint. I do not know whether $f$ gives an isomorphism on stable homotopy groups. $\endgroup$ – Neil Strickland Nov 15 '13 at 9:51
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    $\begingroup$ @Neil: I disagree with your claim that this is a weak homotopy equivalence. $\endgroup$ – André Henriques Nov 15 '13 at 12:45
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    $\begingroup$ Let $Y$ be the subspace $\{1, 1/2, 1/3, 1/4, \ldots, 0\}$ of $\mathbb{R}$, and let $X$ be $\mathbb{N}$ with the discrete topology. There is a continuous map $X \to Y$ that fixes zero and is the map $n \mapsto 1/n$ on the rest. If you take 0 to be the basepoint and use reduced suspension, this map becomes a map from a wedge of circles to the Hawaiian earring, and so is not a $\pi_1$-isomorphism. However, this depends on using reduced suspension with a particular basepoint, and I'm not sure what the stable $\pi_0$ is. (Do the groups even stabilize if the space is not well-pointed?) $\endgroup$ – Tyler Lawson Nov 15 '13 at 13:52
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    $\begingroup$ @Andre: The map in Neil's anwer is indeed a weak homotopy equivalence: The $l$-th homotopy group of $Y$ is the product of all $\pi_l(S^k)$, $k=1,..$, and $\pi_l(S^k)$ is the sum of those. For every $l$, there are only finitely man $k$ such that $\pi_l(S^k) \neq 0$, therefore sum is isomorphic to product. $\endgroup$ – Lennart Meier Nov 15 '13 at 15:35

Let's be precise about the question! I claim it is not meaningful until you choose basepoints in X and Y and restrict to based maps, since otherwise the suspension used to define the stable homotopy groups is ambiguous. And then you might well get different answers for different choices of basepoint: some might be degenerate, others nondegenerate, in the same space. Algebraic topologists tend to define away point-set horrors such as degenerate basepoints. Alternatively, growing a whisker on a bad basepoint gives a good one, and that is actually a cofibrant approximation for the h-model structure ("classical" or "Strom") on based spaces. The answer is yes for based maps between nondegenerately based spaces. I'm not even sure you have a suspension isomorphism for reduced homology if the basepoint is degenerate, but you certainly do if it is nondegenerate (e.g page 107 of "A concise course in algebraic topology"). Now suspend twice to get an isomorphism on homology between simply connected spaces, etc.

  • $\begingroup$ Many thanks! What I care for is the based maps between nondegenerately based spaces. $\endgroup$ – pvigato Nov 16 '13 at 8:17

As Peter points out, stable homotopy groups are usually defined using the reduced suspension, which requires $X$ and $Y$ to be based. Let's talk about both situations.

As in the comment above, let $Y$ be the subspace $\{1,1/2,1/3,1/4,\ldots,0\}$ of $\mathbb{R}$, and let $X$ be $\mathbb{N}$ with the discrete topology. There is a weak equivalence $X \to Y$.

Let's take 0 to be the basepoint and start taking suspensions. This map becomes a map from a countable wedge of spheres to a higher-dimensional Hawaiian earring that features in a famous paper of Barratt-Milnor (you can prove that this is homeomorphic by the standard "bijection from compact to Hausdorff" argument).

The paper "Homotopy and homology groups of the $n$-dimensional Hawaiian earring" by Eda and Kawamura essentially shows that for $n > 1$, the map $X \to Y$, on $\pi_n$, becomes the embedding $\oplus \mathbb{Z} \to \prod \mathbb{Z}$ from a countable direct sum to a countable product. Therefore, the map on stable homotopy groups is not an isomorphism. (This would be some very small manifestation of Tom Goodwillie's result from the comments.)

However, you can also define stable homotopy groups using iterated unreduced suspension (the groups only become well-defined after a couple of suspensions due to basepoint issues). The unreduced suspension is more homotopically well-behaved, and in particular preserves weak equivalence because it only collapses along cofibrations.

  • $\begingroup$ Thank all of you very much! All comments are very helpful and useful for me! Have a good weekend! $\endgroup$ – pvigato Nov 16 '13 at 8:18

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