# Matt N.

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bio website tancast.com/wp-content/… location age member for 3 years, 7 months seen Aug 27 at 6:36 profile views 5,726

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 Aug19 comment The homology groups of $T^2$ by Mayer-Vietoris Thank you so much for your patience!! I understand it so much better now! Aug19 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? Aug19 comment The homology groups of $T^2$ by Mayer-Vietoris Thank you! Where do you use the orientation? Aug19 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.... Aug19 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 $H_1(A \cap B) \cong \mathbb{Z} \oplus \mathbb{Z}$ Aug19 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? Aug19 comment The homology groups of $T^2$ by Mayer-Vietoris and $(i_\ast, j_\ast) (\alpha, \alpha) = (\alpha, \beta)$. Aug19 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 $(i_\ast, j_\ast) (\alpha, 0 ) = (\alpha, 0)$... Aug18 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... Aug18 asked The homology groups of $T^2$ by Mayer-Vietoris Aug18 revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$ deleted 75 characters in body Aug18 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? Aug18 revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$ added 133 characters in body Aug18 comment Follow-up on $H_n(\mathbb{R}^3 - S^1)$ Ryan: I don't know what to make of this. Aug18 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. Aug16 revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$ added 137 characters in body Aug16 revised Follow-up on $H_n(\mathbb{R}^3 - S^1)$ edited body Aug16 comment Homology of $\mathbb{R}^3 - S^1$ Yes, look I posted a follow up question here: math.stackexchange.com/questions/57792/… Aug16 asked Follow-up on $H_n(\mathbb{R}^3 - S^1)$ Aug16 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?