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).

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  • $\begingroup$ 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. $\endgroup$ – Quoka Oct 7 '18 at 20:19
  • $\begingroup$ 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 $\phi\times 1\cup 1\times\psi$. $\endgroup$ – user10354138 Oct 8 '18 at 7:34
  • $\begingroup$ That makes sense (if you write this as an answer I'll accept it). Thank you. $\endgroup$ – Quoka Oct 9 '18 at 17:24
  • $\begingroup$ I've added some details of this discussion to the answer. $\endgroup$ – user10354138 Oct 9 '18 at 18:11

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