Timeline for Why is it true that if two 4-manifolds are homeomorphic then their squares are diffeomorphic?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 20, 2016 at 5:49 | vote | accept | kosta | ||
| Apr 14, 2016 at 21:38 | comment | added | kosta | Thanks for the answer! I think I will need some time to fully comprehend, so please forgive me that I don't accept yet... | |
| Apr 13, 2016 at 15:40 | history | edited | Oscar Randal-Williams | CC BY-SA 3.0 |
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| Apr 13, 2016 at 15:03 | comment | added | Igor Belegradek | This is just a note to myself so I do not forget how the class in $H^3$ arises. When we try to homotope $d$ to a constant map the obstruction to building the homotopy on the $k$-skeleton of $X$ is in $H^k(X;\pi_k(Top/O))$. The only obstruction is in $H^3(X;\pi_3(Top/O))$, and if that obstruction vanishes the homotopy can be build. In particular your argument works if $H_1(X:\mathbb Z_2)=0$. I suspect in Ivan Smith's paper he talks about simply-connected $X$, so your argument applies in this case. | |
| Apr 13, 2016 at 14:46 | comment | added | Oscar Randal-Williams | So it is. I suppose that breaks the argument, unless there is some reason the associated class in $H^3(X;\mathbb{Z}/2)$ must vanish. | |
| Apr 13, 2016 at 14:17 | comment | added | Igor Belegradek | $TOP/O$ is not $6$-connected as $\pi_3(TOP/O)=\mathbb Z_2$. Rather $PL/O$ is $6$-connected and your nice argument works with $TOP$ replaced by $PL$. | |
| Apr 13, 2016 at 12:22 | history | answered | Oscar Randal-Williams | CC BY-SA 3.0 |