Timeline for Homotopy equivalence of certain kinds of adjunction spaces
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| S Nov 22, 2014 at 16:05 | history | suggested | Riccardo | CC BY-SA 3.0 |
corrected last composition, clearly a typo of the writer
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| Nov 22, 2014 at 15:45 | review | Suggested edits | |||
| S Nov 22, 2014 at 16:05 | |||||
| May 3, 2012 at 9:56 | vote | accept | Victor | ||
| May 3, 2012 at 9:56 | vote | accept | Victor | ||
| May 3, 2012 at 9:56 | |||||
| May 3, 2012 at 2:20 | history | edited | Victor | CC BY-SA 3.0 |
Typo in a key formula
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| Apr 27, 2012 at 11:06 | vote | accept | Victor | ||
| May 3, 2012 at 9:56 | |||||
| Apr 27, 2012 at 10:59 | comment | added | Victor | Tom, thinking more about your example I was trying to produce one for which the union $AUB$ is not simply connected. Consider the Hawaiian earring $H$ lying on the $xy$-plane in $\mathbb{R}^3$ in the usual way. Let $A$ be the cone over $H$, with vertex at $(0,0,1)$. Now, rotate $H$ $180$ degrees about the $z$-axis and let $B$ be the cone over this new space, with vertex at $(0,0,-1)$. Both $A$ and $B$ are contractible and have a single point in common. But the fundamental group of the union has at least countably many generators. | |
| Apr 26, 2012 at 17:38 | vote | accept | Victor | ||
| Apr 27, 2012 at 11:06 | |||||
| Apr 26, 2012 at 17:27 | comment | added | Victor | Tom, nice example. Thank you! The pair $(B,(0,0))$ does not have the HEP. It was harder for me to find examples of pairs $(X,C)$ not having the HEP, with $C$ closed. Hatcher gives only one such example in his book (p. 14). Here is another: let $X$ equal the closed subspace of $\mathbf{R}^{2}$ consisting of the graph of $y=sin(1/x)$ for $x\in(0,1]$, the segment $C=0\times[−1,1]$, and an arc connecting $(1,sin(1))$ to the origin. (Exercise 1.3.7 in Hatcher's has a picture of this space). The pair $(X,C)$ does not have the HEP. | |
| Apr 26, 2012 at 16:21 | comment | added | Ronnie Brown | I mention that I came across the gluing theorem in about 1966 by first generalising the result that a homotopy equivalence of spaces induces an isomorphism of homotopy groups. The generalisation replaces the pair $(S^n,x)$ by a pair $(X,A)$ satisfying the HEP. (Section 7.2 of "Topology and groupoids".) I like to have the general result available. | |
| Apr 26, 2012 at 14:47 | history | edited | Victor | CC BY-SA 3.0 |
Corrected bad notation
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| Apr 26, 2012 at 1:33 | comment | added | Tom Goodwillie | For example, in the plane let $A$ consist of the points $(-1/n,0)$ ($n\ge 1$) and $(0,0)$ together with line segments from these points to $(0,-1)$, and let $B$ consist of the points $(1/n,0)$ and $(0,0)$ together with line segments from these points to $(0,1)$. Then $A$ and $B$ are contractible, their intersection is a point, but $A\cup B$ is not contractible. | |
| Apr 26, 2012 at 1:21 | history | edited | Victor | CC BY-SA 3.0 |
Correction of mistake
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| Apr 25, 2012 at 16:32 | comment | added | Tom Goodwillie | In your example $Z\cup_{\phi\circ f}X$ is a (non-Hausdorff) space with two points, but it is of course true that it's not homotopy equivalent to a circle. There are also counterexamples in which everything is compact and Hausdorff. | |
| Apr 25, 2012 at 8:52 | history | answered | Victor | CC BY-SA 3.0 |