Timeline for Counterexamples in algebraic topology?
Current License: CC BY-SA 3.0
23 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 16, 2013 at 23:45 | history | edited | David White | CC BY-SA 3.0 |
Fixed typos, since this was on the front page anyway
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| Mar 29, 2013 at 1:56 | answer | added | Ricardo Andrade | timeline score: 12 | |
| Feb 17, 2011 at 10:24 | history | edited | Johannes Ebert | CC BY-SA 2.5 |
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| Feb 17, 2011 at 8:57 | comment | added | Gil Kalai | So far, most answers were towards the specific 4 questions at the end of the question. If the intention was to create a larger list of counter examples in algebraic topology (or just homotopy theory) perhaps it should be explicitely encouraged... | |
| Feb 14, 2011 at 17:18 | answer | added | Tyler Lawson | timeline score: 15 | |
| Feb 14, 2011 at 11:15 | history | edited | Johannes Ebert | CC BY-SA 2.5 |
another subquestion added
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| Feb 14, 2011 at 10:21 | comment | added | Johannes Ebert | @Elizabeth: this was indeed motivated by a paper I read, where this argument was used to show that $f$ is a homotopy equivalence. In the case under question, the argument was valid since both spaces were simple (and hence $\pi_n X \cong [S^n; X]$). The case of finite CWs is trivial, and in the infinite case I listed a counterexample in my question. Likewise, if $T$ can be infinite, the question is trivial. | |
| Feb 14, 2011 at 10:15 | comment | added | Johannes Ebert | @John: that could be right, Hatcher is using maps $S^{n+m} \to S^n$ which are in the image of the J-homomorphism. If choosen correctlx, these cannot be detected by Steenrod operations. | |
| Feb 14, 2011 at 7:08 | answer | added | Sam Isaacson | timeline score: 28 | |
| Feb 14, 2011 at 6:27 | comment | added | Elizabeth S. Q. Goodman | Can you motivate your question #2 ($[T, X]$ and $[T, Y]$) a little bit more? My first guess is that at least when X or Y is finite, if you just take T=X or Y respectively, you could find a representative map in [X, Y] corresponding to the identity on X or Y, determining a weak equivalence. Maybe you can even extend the argument using skeletons of X or Y. So I don't know a counterexample to that. | |
| Feb 14, 2011 at 5:59 | answer | added | Gil Kalai | timeline score: 16 | |
| Feb 14, 2011 at 5:40 | history | made wiki | Post Made Community Wiki by S. Carnahan♦ | ||
| Feb 14, 2011 at 3:48 | history | edited | Allen Knutson | CC BY-SA 2.5 |
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| Feb 14, 2011 at 3:46 | comment | added | Tom Goodwillie | For #3: $X$ could be empty. Also, slightly less silly remark, without the finiteness assumption there is the Eilenberg swindle: Let $X$ be the product of infinitely many copies of $Y$ and let $Z$ be a point. | |
| Feb 14, 2011 at 1:52 | comment | added | Sam Isaacson | For #1, what about a $\mathbf{Z}/4$ Moore space versus a $\mathbf{Z}/8$ Moore space? | |
| Feb 14, 2011 at 1:51 | answer | added | Eric Wofsey | timeline score: 13 | |
| Feb 14, 2011 at 1:44 | comment | added | John Klein | @Johannes: I haven't checked it would wager a eurocent that Hatcher's examples in mathoverflow.net/questions/5431 do what you want in #1: they are total spaces of spherical fibrations over spheres that carry a section. | |
| Feb 14, 2011 at 1:44 | answer | added | Jeff Strom | timeline score: 12 | |
| Feb 14, 2011 at 1:39 | comment | added | John Klein | @John Palmieri: The suspension of $S^1 \times S^1$ is homotopy equivalent to $S^2 \vee S^2 \vee S^3$, because the suspension of the attaching map (the Whitehead product) is null homotopic. @Johannes: If $X = S^p \times S^q$ and $Y = S^p \vee S^q \vee S^{p+q}$, then $H^*(X)$ and $H^*(Y)$ are isomorphic as modules over the Steenrod algebra (however, the ring structures are not isomorphic). | |
| Feb 14, 2011 at 0:34 | comment | added | Johannes Ebert | For #1: these spaces are homotopy equivalent, see Hatcher Proposition 4I.1. For #3: the question is indeed pretty trivial if the spaces can be of infinite type. You can also use that $\mathbb{R}^2 \cong \mathbb{R}$ as abelian groups. | |
| Feb 14, 2011 at 0:24 | comment | added | John Palmieri | For #1, what about a suspension of a torus versus $S^2 \vee S^2 \vee S^3$? For #3, it's pretty easy to find examples where $X$ is not a finite CW complex (say, an infinite product of $Y \times Z$). | |
| Feb 14, 2011 at 0:14 | history | asked | Johannes Ebert | CC BY-SA 2.5 |