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 $\phi_{1} \times \psi_{1}$?


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.

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  • $\begingroup$ Thank you Lee. This answer was so helpful. Unfortunately this map $\gamma_{\alpha,\beta}(x,y)$ doesn't give you much intuition how to build the product of two complexes from knowing how to build $X$ and $Y$. Also, if you want I can tag you in the answers that said the the attachment maps are the product of the attachment maps. $\endgroup$ – user404735 Apr 26 at 4:39
  • $\begingroup$ Well, I don't know about intuition, but the formula certainly gives you precise instructions. To get intuition, you could try constructing several examples using the formula. Perhaps the intuition comes in learning to visualize examples, starting with $D^1 \times D^1$ (the square), and $D^2 \times D^1$ (the cylinder). More complex examples tend to not be visualizable in ordinary $\mathbb R^3$ for different reasons, namely that most such examples do not embed in $\mathbb R^3$. $\endgroup$ – Lee Mosher Apr 26 at 13:37
  • $\begingroup$ For example, try to visualize $Y_1 \times Y_2$ where both $Y_1$ and $Y_2$ are homeomorphic to "$Y$"; it's a challenge because this is a 2-dimensional complex (easy, right?) which is not embeddable in $\mathbb R^3$ (okay, not so easy...). $\endgroup$ – Lee Mosher Apr 26 at 13:40
  • $\begingroup$ Yeah but I was hoping these specific instructions would help me easily calculate the product of chain complexes. For example, I can easily compute the cellular homology of the torus using the cell structure I described in my question. But say I wanted to compute the cellular homology of a torus using your map. Do these specific instructions describe the boundary operator in the cellular chain complex of a torus? I don't see how it does. $\endgroup$ – user404735 Apr 26 at 23:07
  • $\begingroup$ Yes it does. But, that's another question. $\endgroup$ – Lee Mosher Apr 26 at 23:52

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