Cell Complex of Cartesian Product Let $X$ and $Y$ be cell complexes. Then $X \times Y$ has the structure of a cell complex with cells the product of $e^{m}_{\alpha} \times e^{n}_{\beta}$ where $e^{m}_{\alpha}$ ranges over the cells of $X$ and $e^{n}_{\beta}$ ranges over the cells of $Y$. So firstly, I have seen many answers that say the attachment map for the cell $e^{m}_{\alpha} \times e^{n}_{\beta}$ is the corresponding map is $\phi_{\alpha} \times \psi_{\beta}$ where $\phi_{\alpha}$ is the attachment map of the boundary of $e^{m}_{\alpha}$ and $\psi_{\beta}$ is the attachment map of $e^{n}_\beta$. I don't think I understand the product map. Here are my issues.
1) If $e^{m}_{\alpha} \times e^{n}_{\beta}$ is a cell in the product then we must glue a disk of dimension $n+m$. Is it true that we know how to glue a disk $D^{n+m+}$ because it is homemorphic to $D^{m} \times D^{n}$ and we know how to attach the products of cells?
My main question is I really don't understand how this attachment map works. 
I want to bring up an example to show where I am confused. The torus is a cell complex with 1 o-cell, then attaching two 1- cells. Then we attach a $2- cell$ along the wedge of circles following the path $aba^{-1}b^{-1}$. Now lets try to build the torus from knowing how to build $S^{1}$. For one $S^{1}$ let $e_{0}$ be the 0-cell and $e_{1}$ be the  1-cell with attachment map $\phi_{1}$. For the second circle denote $f_{0}$ as the o -cell and $f_{1}$ as the 1-cell with attachment map $\psi_{1}$. Lets now build the torus. So our only 0-cell is $e_{0} \times f_{0}$. So say we are at the stage with a wedge of two circles. How do we attach the cell $e_{1} \times f_{1}$ using $\phi_{1} \times \psi_{1}$? 
 A: I don't know why someone would say that the attaching map of $e_m \times e_n$ has the form $\phi_\alpha \times \psi_\beta$. That would seem to imply that the domain of the attaching map is $S^{m-1} \times S^{n-1}$, which does not even have the right dimension for the boundary of a cell of dimension $m+n$.
Instead the boundary of $D^m \times D^n$  is $(S^{m-1} \times D^n) \cup (D^m \times S^{n-1})$. You can convince yourself that this is homeomorphic to $S^{m+n-1}$, via the restriction of some homeomorphism from $D^m \times D^n$ to $D^{m+n}$.
So the attaching map for $e^m_\alpha \times e^n_\beta$ must be a function of the form 
$$\gamma_{\alpha,\beta} : (S^{m-1} \times D^n) \cup (D^m \times S^{n-1}) \to (X \times Y)^{(m+n-1)}
$$
We already have attaching maps for the cells $e^m_\alpha$ and $e^n_\beta$  of the form
$$\phi_\alpha : S^{m-1} \to X^{(m-1)}
\qquad\qquad
\psi_\beta : S^{n-1} \to Y^{(n-1)}
$$
which extend to characteristic maps for those cells of the form
$$\chi_\alpha : D^m \to X^{(m)} \qquad\qquad \omega_\beta : D^n \to Y^{(n)}
$$
The definition of the attaching map for $e^m_\alpha \times e^n_\beta$ can therefore be given by the function
$$\gamma_{\alpha,\beta}(x,y) = \begin{cases}
(\phi_\alpha(x),\omega_\beta(y)) & \quad\text{if $(x,y) \in S^{m-1} \times D^n$} \\
(\chi_\alpha(x),\psi_\beta(y)) &\quad\text{if $(x,y) \in D^m \times S^{n-1}$}
\end{cases}
$$
and one should note that 
$$(\phi_\alpha(x),\omega_\beta(y)) \in X^{(m-1)} \times Y^n \subset (X \times Y)^{m+n-1}
$$ 
and that 
$$(\chi_\alpha(x),\psi_\beta(y)) \in X^m \times Y^{n-1} \subset (X \times Y)^{m+n-1}
$$
Seeing the formula for $\gamma_{\alpha,\beta}$, one could say that $\gamma_{\alpha,\beta}$ is the restriction to $(S^{m-1} \times D^n) \cup (D^m \times S^{n-1})$ of the product of the characteristic maps $\chi_\alpha \times \omega_\beta$.
But it is certainly wrong to say it is the product of the attaching maps $\phi_\alpha \times \psi_\beta$. I'm curious to know where you saw such answers.
