Let $$ H= \left\{ \begin{pmatrix} 1 & z_1 & z_3 \\ 0 & 1 & z_2 \\ 0 & 0 & 1 \end{pmatrix} : z_1,z_2,z_3 \in \mathbb{C} \right\} $$ be the complex Heisenberg group and denote by $G$ the subgroup of $H$ where the entries $z_1,z_2,z_3 \in \mathbb{Z}[i]$ are in the Gaussian integers. Then $G$ acts on $H$ from the left and the quotient $W=H/G$ is a compact manifold with 6 real dimensions (3 complex dimensions, but $W$ is not Kähler). $W$ is called Iwasawa manifold.
Note that $$ \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & z_1 & z_3 \\ 0 & 1 & z_2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & z_1+a & z_3+a z_2+c \\ 0 & 1 & z_2+b \\ 0 & 0 & 1 \end{pmatrix}, $$ i.e. every element in $W$ has exactly one representative with elements $z_1,z_2,z_3 \in [0,1)\times [0,1)i$. It is mentioned e.g. in Daniel Huybrecht: Complex Geometry that $W$ is a non-trivial $2$-torus bundle over a $4$-torus (the non-triviality is checked by computing homology groups).
Question 1: Consider the map $$\Phi: ([0,1]/_{\{0,1\}} \times [0,1]/_{\{0,1\}} i)^3 \rightarrow W, (z_1,z_2,z_3) \mapsto \begin{pmatrix} 1 & z_1 & z_3 \\ 0 & 1 & z_2 \\ 0 & 0 & 1 \end{pmatrix}.$$ This is a well-defined (i.e. $\Phi(0,0,0)=\Phi(1,0,0)=\Phi(0,1,0)=\dots$), bijective (statement about representatives for elements in $W$ above), continuous map. Thus it is a homeomorphism. However it can't be, because it has different homology groups. What is the problem with my "homeomorphism"?
And
Question 2: What is a good way to define functions on $W$? I was hoping to use above bijection, but there is something wrong with it, so it seems like a bad idea.