Timeline for Homology of a closed $3$-manifold with balls removed
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2021 at 10:12 | comment | added | Piotr | Alternatively, one could quotient M by a complement of a collar neighborhood of B to get a degree-1 map $F:M\to \bigvee S^3$ and then use naturality of MV to finish the computation, since MV sequence for $S^3$ is, essentially, unique by algebra. | |
| Feb 19, 2021 at 10:11 | comment | added | Piotr | Also, I think any construction of MV that satisfies a bunch of axioms has to give the same maps as the construction coming from axioms - but I don't know if anybody has written it down. | |
| Feb 19, 2021 at 10:10 | comment | added | Piotr | Mike, I think Eduardo is not sure how to prove that your construction of the boundary map is a correct one. @EduardoLonga , check any proof of Mayer-Vietoris which uses singular homology to see that one can construct a MV sequence this way. | |
| Dec 8, 2020 at 19:30 | comment | added | mme | It is perhaps not helpful to delete all the balls at once. If you delete a single ball, you find that $H_2(M \setminus B) = H_2(M)$; the cost of deleting $B$ is that the third homology dies. After that, every new ball we delete adds to second homology, and in fact, if $X$ is a noncompact $n$-manifold and $D \subset X$ is an n-disc, then $X \setminus D^\circ \simeq X \vee S^{n-1}$. I do not really want to write down the proof, though. To see this, when you are in the process of contracting $X$ to a complex one dimension lower, you should delete the last n-cell you contract. | |
| Dec 8, 2020 at 19:28 | comment | added | mme | I am not sure what confusion remains. Are you worried about orientations? Ignoring orientations it should be clear that $\partial(M \setminus B)$ is precisely the union of spheres, and the simplices appearing in the boundary operator are precisely the simplices in the triangulation of that sphere. | |
| Dec 8, 2020 at 19:11 | comment | added | Eduardo Longa | @MikeMiller Yes, this is exactly what I have to prove. Is it obvious? | |
| Dec 8, 2020 at 19:08 | comment | added | mme | Doing this explicitly (break it up into the fundamental chain for $M \setminus B$ and the fundamental chain for $B$) sends the fundamental class of $M$ to the sum of the fundamental classes of the spheres. | |
| Dec 8, 2020 at 19:08 | comment | added | mme | Your guess follows from an explicit description of these groups and their boundary maps. $H_3(M)$ is generated by the fundamental class, a sum of oriented simplices. $H_2(U \cap V) \simeq \sqcup_n S^2$ is generated by the $n$ fundamental classes of these distinct spheres. And the boundary map is obtained by breaking up a chain on $M$ as the sum of a chain in $U$ and a chain in $V$, taking the boundary, and seeing what class in $U \cap V$ gives that boundary. . | |
| Dec 8, 2020 at 18:47 | history | asked | Eduardo Longa | CC BY-SA 4.0 |