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seen Aug 27 at 6:36

Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
Thank you so much for your patience!! I understand it so much better now!
Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
And where does $0 \rightarrow Coker f \rightarrow H_1(T) \rightarrow Ker g \rightarrow 0$ come from?
Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
Thank you! Where do you use the orientation?
Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
Can I ask you one more question: Dylan mentioned orientation. Where does that come in? I think neither of you uses it anywhere....
Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
OK, I think I start to understand: you can write the map as this matrix because you write an element $\alpha$ in $H_n(A\cap B)$ as $(\alpha, 0)$ even though the elements in $H_1(A\cap B)$ are not actually pairs. Well they kind of are because $H_1(A \cap B) \cong \mathbb{Z} \oplus \mathbb{Z}$
Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
Thank you! When you wrote cocycles you meant to write cycles really, didn't you?
Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
and $(i_\ast, j_\ast) (\alpha, \alpha) = (\alpha, \beta)$.
Aug
19
comment The homology groups of $T^2$ by Mayer-Vietoris
Very nice, thank you! I understand that $(i_\ast, j_\ast) (\alpha, 0 ) = (i_\ast, j_\ast) (\beta, 0 )$ but how is $(i_\ast, j_\ast) (\alpha, 0 ) = (\alpha, \beta)$? I think $(i_\ast, j_\ast) (\alpha, 0 ) = (\alpha, 0)$...
Aug
18
comment Follow-up on $H_n(\mathbb{R}^3 - S^1)$
@Dylan: Yes but $A \cap B \cong \mathbb{Z}$ so I don't even have a surjective function. And besides I want it to be the zero function because I want an isomorphism further down the chain...
Aug
18
asked The homology groups of $T^2$ by Mayer-Vietoris
Aug
18
revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$
deleted 75 characters in body
Aug
18
comment Follow-up on $H_n(\mathbb{R}^3 - S^1)$
@Dylan: and shouldn't I be looking at $H_1(A \cap B) \rightarrow H_1(A) \oplus H_1(B)$ instead?
Aug
18
revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$
added 133 characters in body
Aug
18
comment Follow-up on $H_n(\mathbb{R}^3 - S^1)$
Ryan: I don't know what to make of this.
Aug
18
comment Follow-up on $H_n(\mathbb{R}^3 - S^1)$
@Dylan: the two cycles that generate $H_1(T^2)$ are the one that goes around the centre hole and the one that is perpendicular to it. Only the first is in $A \cap B$. From this I can conclude that $f: H_1(A \cap B) \rightarrow H_1(A)$ is not surjective.
Aug
16
revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$
added 137 characters in body
Aug
16
revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$
edited body
Aug
16
comment Homology of $\mathbb{R}^3 - S^1$
Yes, look I posted a follow up question here: math.stackexchange.com/questions/57792/…
Aug
16
asked Follow-up on $H_n(\mathbb{R}^3 - S^1)$
Aug
16
comment Homology of $\mathbb{R}^3 - S^1$
How do you compute $H_1$ and $H_2$? I'm stuck : ( Could you expand on your answer please?