Timeline for Is the Euler characteristic of aspherical connected 2-complexes at most 1? (No!) What can be said about subcomplexes of 2-complexes deformation retractible onto graphs.
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| Feb 6, 2012 at 2:08 | comment | added | Arnaud Mortier | @Jesper Gordal : here we assume that $\chi(B)=1$. If you still know such examples which are not contractible, then it is an answer to Alexey's question. @Alexey : Still thinking of a counterexample, put $X=\mathbb{T}^2\setminus\mathbb{D}^2$, a torus minus a little disc, and $X_0:=X$. Then to each of the two obvious generators of $\pi_1(X0)$, glue a copy of $X$ along its boundary : this forms $X_1$. Repeating this process we obtain an inductive limit which is a CW-complex with every chance to be a counterexample. | |
| Feb 2, 2012 at 15:22 | comment | added | Alexey Muranov | It has been explained to me by a colleague why $B$ will be acyclic: $H_1(B)$ with coefficients in any $\mathbb Z_p$ has to be trivial, not only with coefficients in $\mathbb Q$. | |
| Feb 1, 2012 at 11:59 | comment | added | Alexey Muranov | Sorry about the silly question: of course, if it is acyclic but $\pi_1$ is nontrivial, it is not contractible. I do not see though an easy example. | |
| Feb 1, 2012 at 9:12 | comment | added | Alexey Muranov | @Jesper Grodal: can you point me to an example of an acyclic non contractible 2-complex, please? By the way, i do not see why the subcomplex $B$ will be acyclic, i only see that $H_1(B)$ will be finite. | |
| Jan 31, 2012 at 15:09 | comment | added | Jesper Grodal | Unfortunately there exists acyclic 2-complexes which are not contractible, so this approach won't work. | |
| Jan 31, 2012 at 12:16 | comment | added | Alexey Muranov | Hi Arnaud, what is $D\pi_1(X)$? | |
| Jan 31, 2012 at 12:07 | history | answered | Arnaud Mortier | CC BY-SA 3.0 |