If $\mathbb{Z}$ acts on $\mathbb{C}^n \backslash \{0\}$ by $(m,z) \mapsto 2^m\,z$, I need to show that $H^2_{DR}(X=(\mathbb{C}^n \backslash \{0\})/\mathbb{Z},\mathbb{C})=0$.
I start with showing it is a complex manifold. This is a little weak for me so if you can fill in that would be great! I know that the quotient space inherits its complex structure from $\mathbb{C}^n$, and the action of the group is properly discontinuous and free, and so the quotient space is a complex manifold. I would like to see this from a more fundamental level, i.e., transition functions and coverings.(?)
Moving forward I then show $H^2_{DR}=0$ by showing that $(\mathbb{C}^n \backslash \{0\})/\mathbb{Z}$ is diffeomorphic to $S^{2n-1} \times S^1$. In my class notes, it is mentioned that the composite $S^{2n-1} \times [1/2,1] \to \mathbb{C}^n\backslash \{0\}$ given by $(v,t) \mapsto t\,v$. Somehow this induces $S^{2n-1}\times S^1 \cong X$. This I could use some explanations.(?) I believe that since the rank of the differential of this map is full, that this map is a diffeomorphism?
Lastly, using Kunneth's formula, I have $H^2_{DR}(X,\mathbb{C}) \cong H^0_{DR}(S^{2n-1},\mathbb{C}) \otimes H^2_{DR}(S^1,\mathbb{C})\\ \oplus H^1_{DR}(S^{2n-1}) \otimes H^1_{DR}(S^1)\\ \oplus H^2_{DR}(S^{2n-1},\mathbb{C}) \otimes H^0_{DR}(S^1,\mathbb{C})\\ \cong 0 + H^1_{DR}(S^{2n-1}) \otimes H^1_{DR}(S^1) + 0 \\ \cong H^1_{DR}(S^{2n-1}) \otimes H^1_{DR}(S^1)$. But as $H^1$ tells of the "holes" in something, I don't see how either of these are zero. Can someone point out my mistake?