# Attaching maps of products of CW-complexes.

I am a bit confused about how the "product attaching-maps" work.

For instance, if I wanted to find the cell structure for $$S^1\times S^1$$ then I can proceed similarly to here.

So we will call $$e_0$$ and $$f_0$$ the zero-cells of $$S^1$$ and $$S^1$$, respectively. Similarly, $$e_1$$ and $$f_1$$ are the respective one-cells. Then $$\phi$$ and $$\psi$$ are the attaching maps taking the boundary of $$D^1$$ to $$e_0$$ and $$f_0$$ respectively.

When we attach the cell $$e_1\times f_1$$ to the one-skeleton of $$S^1\times S^1$$, we do so via the attaching map $$\phi\times \psi$$. But $$\phi\times \psi$$ are only defined on $$\partial D^1 \times \partial D^1$$, and not on $$\partial (D^1\times D^1) = (\partial D_1\times D_1) \cup (D_1\times \partial D_1)$$. This is where I get confused.

If anyone can explain what's going on I would highly appreciate it.

Be careful! You forgot the 1-cells $$e_0\times f_1$$ and $$e_1\times f_0$$ which are attached to the 0-cell $$e_0\times f_0$$ in the obvious way to form the 1-skeleton $$S^1\vee S^1$$ of $$S^1\times S^1$$. Now glue your 2-cell $$e_1\times f_1$$ to the 1-skeleton.
Edit to bring the discussion into the answer The attaching map $$\phi$$ of a cell $$e^n$$ of a CW-complex $$X$$ defines the characteristic map of this $$n$$-cell $$e^n\hookrightarrow X_{n-1}\amalg e^n\to X_n\hookrightarrow X$$ and restricting the characteristic map to $$\partial e^n$$ gives the attaching map, so with an abuse of notation they have the same name $$\phi$$. Then the product of attaching maps is actually taken as the restriction of the set-theoretic product of characteristic maps (using $$e^n\times e^m\cong e^{n+m}$$), so they are not the set-theoretic product of attaching maps but $$\phi\times 1\cup 1\times\psi$$ (where the $$1$$ is the obvious identification).
• Sorry I didn't phrase what I mean properly. I see that we have a zero-cell and two-one cells and their attachment maps make sense to me. My issue is that when I glue the 2-cell to the one-skeleton, the attaching map is not well-defined. Intuitively, I see what should happen. My issue is with where I am misunderstanding how $\phi\times\psi$ works. – Quoka Oct 7 '18 at 20:19
• The attaching map is well-defined. The attaching map that we write as $\phi\times\psi$ is not the set-theoretic $\phi\times\psi$, but $\phi\times 1\cup 1\times\psi$. – user10354138 Oct 8 '18 at 7:34