Timeline for Homotopy domination of a wedge of two polyhedra
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2021 at 16:25 | comment | added | Tyrone | @NicholasKuhn $\alpha\circ 5$ and $5\circ\alpha$ are the same in $\pi_6S^3$ by an Eckmann-Hilton argument. Normally I would understand $5\cdot\alpha=\alpha\circ 5$ for an element $\alpha\in\pi_nZ$. It seems to me that $X,Y$ will be actually be equivalent when localised at any prime. | |
| Mar 25, 2021 at 21:52 | comment | added | Nicholas Kuhn | I think this fits with my old comments as follows: $X$ and $Y$ will be equivalent localized away from 5 and also completed at 5 (when both will just be $S^3 \vee S^7$). By the way, does $\beta = 5\alpha$ mean $\alpha \circ 5$ or $5 \circ \alpha$ in $\pi_6(S^3)$? (Are these the same?) | |
| Mar 25, 2021 at 17:52 | comment | added | Tyrone | @BenWieland $X$ and $Y$ are not homotopy equivalent. Assume a homotopy equivalence $X\rightarrow Y$ and look at the induced maps between the Puppe sequences of the two complexes. You get degree $\pm1$ maps on $S^7$ and on $S^4=\Sigma S^3$. But no choice of signs is compatible with the suspensions of $\alpha,\beta$. | |
| Mar 25, 2021 at 17:00 | comment | added | Ben Wieland | Shearing is an interesting idea, but... Aren't $X$ and $Y$ homotopy equivalent? Isn't the key what subgroup the attaching map generates? | |
| Mar 24, 2021 at 21:05 | comment | added | Tyrone | @IJL corrected. | |
| Mar 24, 2021 at 21:04 | history | edited | Tyrone | CC BY-SA 4.0 |
edited body
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| Mar 24, 2021 at 20:55 | comment | added | IJL | In line 4, two occurences of $S^7$ should be $e^7$; I lack the authority to edit. | |
| Mar 24, 2021 at 19:07 | history | edited | Tyrone | CC BY-SA 4.0 |
Three years later I perhaps understand the question.
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| Nov 13, 2017 at 16:34 | comment | added | M.Ramana | Thank you for your answer. Clearly, $\mathbb{S}^n$ is homotopy dominated by $\mathbb{S}^n \vee \mathbb{S}^n$ (even is a retract of $\mathbb{S}^n \vee \mathbb{S}^n$) . But if $x$ is the wedge point of $\mathbb{S}^n \vee \mathbb{S}^n$, then we can consider $\mathbb{S}^n$ itselt as the form $\mathbb{S}^n \vee \{ x\}$. | |
| Nov 13, 2017 at 16:13 | history | answered | Tyrone | CC BY-SA 3.0 |