Timeline for Must a weak homotopy equivalence induce an isomorphism between stable homotopy groups?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Nov 16, 2013 at 8:13 | vote | accept | pvigato | ||
| Nov 16, 2013 at 5:00 | answer | added | Tyler Lawson | timeline score: 10 | |
| Nov 16, 2013 at 2:02 | answer | added | Peter May | timeline score: 11 | |
| Nov 15, 2013 at 20:37 | comment | added | André Henriques | Omar and Lennart: you're right, I'm wrong. | |
| Nov 15, 2013 at 18:13 | comment | added | Omar Antolín-Camarena | Jinx, you owe me a coke, @LennartMeier! | |
| Nov 15, 2013 at 17:09 | comment | added | Tom Goodwillie | (I haven't thought it through fully, but what I have in mind is adapting a Pontryagin-style framed-bordism proof of the Freudenthal suspension theorem.) | |
| Nov 15, 2013 at 17:07 | comment | added | Tom Goodwillie | @Tyler: I think that in your example the groups do stabilize as usual, and in fact that for roughly $j>k$ the reduced $j$th suspension has the same $\pi_{j+k}$ as the infinite product of copies of $S^j$. | |
| Nov 15, 2013 at 15:35 | comment | added | Lennart Meier | @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. | |
| Nov 15, 2013 at 15:33 | comment | added | Omar Antolín-Camarena | Why do you disagree @AndréHenriques? Aren't $pi_n(Y)$ and $\pi_n(X)$ respectively the direct product and the direct sum of all the $\{ \pi_n(S^m) : m \ge 0\}$? Those agree because only $n$ of the groups can be non-zero. | |
| Nov 15, 2013 at 14:59 | history | edited | Fernando Muro |
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| Nov 15, 2013 at 13:52 | comment | added | Tyler Lawson | 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?) | |
| Nov 15, 2013 at 12:45 | comment | added | André Henriques | @Neil: I disagree with your claim that this is a weak homotopy equivalence. | |
| Nov 15, 2013 at 10:16 | history | edited | pvigato | CC BY-SA 3.0 |
edited title
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| Nov 15, 2013 at 9:51 | comment | added | Neil Strickland | 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. | |
| Nov 15, 2013 at 9:36 | review | First posts | |||
| Nov 15, 2013 at 9:38 | |||||
| Nov 15, 2013 at 9:34 | comment | added | Qiaochu Yuan | "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. | |
| Nov 15, 2013 at 9:25 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
fixed latex
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| Nov 15, 2013 at 9:18 | history | asked | pvigato | CC BY-SA 3.0 |