Mixing
Reduced homology group $H_k(S^4 - N^4, mathbb Z)=H^{4-k}(S^4 - N^4,mathbb Z)=H_{k-1}(N^4, mathbb Z)=mathbb...
$begingroup$
Let $N^4$ be a 4-dimensional $D^2 times T^2 = D^2 times S^1 times S^1$.
(let us denote $tilde H$ as the reduced homology or homology group)
I know that
$$
tilde H_0(N^4,mathbb{Z})=0,
$$
$$
H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_2(N^4,mathbb Z)=mathbb Z,
$$
$$
H_3(N^4,mathbb Z)=0
$$
Am I correct that:
$$
tilde H_0(S^4 - N^4,mathbb Z)=0,
$$
$$
H_1(S^4 - N^4, mathbb Z)=mathbb Z,
$$
$$
H_2(S^4 - N^4,mathbb Z)=H^2(S^4 - N^4,mathbb Z)=H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_3(S^4 - N^4,mathbb Z)=0,
$$
However, I obtained that
$$
H_1(S^4 - N^4,mathbb Z)=H^3(S^4 - N^4,mathbb Z)= tilde H_0(N^4, mathbb Z)=0
$$ via Alexander
duality and Poincare duality.
$$
H_3(S^4 - N^4,mathbb Z)=H^1(S^4 - N^4,mathbb Z)= H_2(N^4, mathbb Z)=mathbb Z$$
via Alexander duality
and Poincare duality.
Something is wrong? Could you correct my mistakes? Which part leads to the contradiction of my results?
Thank you!
general-topology algebraic-topology homology-cohomology geometric-topology surgery-theory
asked Jan 30 at 4:11
annie heartannie heart
668721
$endgroup$
add a comment |
$begingroup$
Let $N^4$ be a 4-dimensional $D^2 times T^2 = D^2 times S^1 times S^1$.
(let us denote $tilde H$ as the reduced homology or homology group)
I know that
$$
tilde H_0(N^4,mathbb{Z})=0,
$$
$$
H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_2(N^4,mathbb Z)=mathbb Z,
$$
$$
H_3(N^4,mathbb Z)=0
$$
Am I correct that:
$$
tilde H_0(S^4 - N^4,mathbb Z)=0,
$$
$$
H_1(S^4 - N^4, mathbb Z)=mathbb Z,
$$
$$
H_2(S^4 - N^4,mathbb Z)=H^2(S^4 - N^4,mathbb Z)=H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_3(S^4 - N^4,mathbb Z)=0,
$$
However, I obtained that
$$
H_1(S^4 - N^4,mathbb Z)=H^3(S^4 - N^4,mathbb Z)= tilde H_0(N^4, mathbb Z)=0
$$ via Alexander
duality and Poincare duality.
$$
H_3(S^4 - N^4,mathbb Z)=H^1(S^4 - N^4,mathbb Z)= H_2(N^4, mathbb Z)=mathbb Z$$
via Alexander duality
and Poincare duality.
Something is wrong? Could you correct my mistakes? Which part leads to the contradiction of my results?
Thank you!
general-topology algebraic-topology homology-cohomology geometric-topology surgery-theory
asked Jan 30 at 4:11
annie heartannie heart
668721
$endgroup$
2
$begingroup$
Poincare duality only works on closed manifolds.
$endgroup$
– Cheerful Parsnip
Jan 30 at 4:25
$begingroup$
So how should I derive the above?
$endgroup$
– annie heart
Jan 30 at 4:27
$begingroup$
Have you looked at Hatcher's Corollary 3.45 on pg 255 of Algebraic Toplogy? It is the statement of Alexander duality.
$endgroup$
– Tyrone
Jan 30 at 10:16
add a comment |
$begingroup$
Let $N^4$ be a 4-dimensional $D^2 times T^2 = D^2 times S^1 times S^1$.
(let us denote $tilde H$ as the reduced homology or homology group)
I know that
$$
tilde H_0(N^4,mathbb{Z})=0,
$$
$$
H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_2(N^4,mathbb Z)=mathbb Z,
$$
$$
H_3(N^4,mathbb Z)=0
$$
Am I correct that:
$$
tilde H_0(S^4 - N^4,mathbb Z)=0,
$$
$$
H_1(S^4 - N^4, mathbb Z)=mathbb Z,
$$
$$
H_2(S^4 - N^4,mathbb Z)=H^2(S^4 - N^4,mathbb Z)=H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_3(S^4 - N^4,mathbb Z)=0,
$$
However, I obtained that
$$
H_1(S^4 - N^4,mathbb Z)=H^3(S^4 - N^4,mathbb Z)= tilde H_0(N^4, mathbb Z)=0
$$ via Alexander
duality and Poincare duality.
$$
H_3(S^4 - N^4,mathbb Z)=H^1(S^4 - N^4,mathbb Z)= H_2(N^4, mathbb Z)=mathbb Z$$
via Alexander duality
and Poincare duality.
Something is wrong? Could you correct my mistakes? Which part leads to the contradiction of my results?
Thank you!
general-topology algebraic-topology homology-cohomology geometric-topology surgery-theory
asked Jan 30 at 4:11
annie heartannie heart
668721
$endgroup$
Let $N^4$ be a 4-dimensional $D^2 times T^2 = D^2 times S^1 times S^1$.
(let us denote $tilde H$ as the reduced homology or homology group)
I know that
$$
tilde H_0(N^4,mathbb{Z})=0,
$$
$$
H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_2(N^4,mathbb Z)=mathbb Z,
$$
$$
H_3(N^4,mathbb Z)=0
$$
Am I correct that:
$$
tilde H_0(S^4 - N^4,mathbb Z)=0,
$$
$$
H_1(S^4 - N^4, mathbb Z)=mathbb Z,
$$
$$
H_2(S^4 - N^4,mathbb Z)=H^2(S^4 - N^4,mathbb Z)=H_1(N^4,mathbb Z)=mathbb Z^2,
$$
$$
H_3(S^4 - N^4,mathbb Z)=0,
$$
However, I obtained that
$$
H_1(S^4 - N^4,mathbb Z)=H^3(S^4 - N^4,mathbb Z)= tilde H_0(N^4, mathbb Z)=0
$$ via Alexander
duality and Poincare duality.
$$
H_3(S^4 - N^4,mathbb Z)=H^1(S^4 - N^4,mathbb Z)= H_2(N^4, mathbb Z)=mathbb Z$$
via Alexander duality
and Poincare duality.
Something is wrong? Could you correct my mistakes? Which part leads to the contradiction of my results?
Thank you!
general-topology algebraic-topology homology-cohomology geometric-topology surgery-theory
general-topology algebraic-topology homology-cohomology geometric-topology surgery-theory
asked Jan 30 at 4:11
annie heartannie heart
668721
asked Jan 30 at 4:11
annie heartannie heart
668721
edited Jan 30 at 4:21
annie heart
asked Jan 30 at 4:11
annie heartannie heart
668721
asked Jan 30 at 4:11
annie heartannie heart
668721
asked Jan 30 at 4:11
annie heartannie heart
668721
668721
2
$begingroup$
Poincare duality only works on closed manifolds.
$endgroup$
– Cheerful Parsnip
Jan 30 at 4:25
$begingroup$
So how should I derive the above?
$endgroup$
– annie heart
Jan 30 at 4:27
$begingroup$
Have you looked at Hatcher's Corollary 3.45 on pg 255 of Algebraic Toplogy? It is the statement of Alexander duality.
$endgroup$
– Tyrone
Jan 30 at 10:16
add a comment |
2
$begingroup$
Poincare duality only works on closed manifolds.
$endgroup$
– Cheerful Parsnip
Jan 30 at 4:25
$begingroup$
So how should I derive the above?
$endgroup$
– annie heart
Jan 30 at 4:27
$begingroup$
Have you looked at Hatcher's Corollary 3.45 on pg 255 of Algebraic Toplogy? It is the statement of Alexander duality.
$endgroup$
– Tyrone
Jan 30 at 10:16
2
2
$begingroup$
Poincare duality only works on closed manifolds.
$endgroup$
– Cheerful Parsnip
Jan 30 at 4:25
$begingroup$
Poincare duality only works on closed manifolds.
$endgroup$
– Cheerful Parsnip
Jan 30 at 4:25
$begingroup$
So how should I derive the above?
$endgroup$
– annie heart
Jan 30 at 4:27
$begingroup$
So how should I derive the above?
$endgroup$
– annie heart
Jan 30 at 4:27
$begingroup$
Have you looked at Hatcher's Corollary 3.45 on pg 255 of Algebraic Toplogy? It is the statement of Alexander duality.
$endgroup$
– Tyrone
Jan 30 at 10:16
$begingroup$
Have you looked at Hatcher's Corollary 3.45 on pg 255 of Algebraic Toplogy? It is the statement of Alexander duality.
$endgroup$
– Tyrone
Jan 30 at 10:16
add a comment |
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2
$begingroup$
Poincare duality only works on closed manifolds.
$endgroup$
– Cheerful Parsnip
Jan 30 at 4:25
$begingroup$
So how should I derive the above?
$endgroup$
– annie heart
Jan 30 at 4:27
$begingroup$
Have you looked at Hatcher's Corollary 3.45 on pg 255 of Algebraic Toplogy? It is the statement of Alexander duality.
$endgroup$
– Tyrone
Jan 30 at 10:16