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!