3
$\begingroup$

For a CW-complex, there's the cellular boundary formula that $$ d_n(e^n_\alpha)=\sum_\beta d_{\alpha\beta}e^{n-1}_\beta $$ where the coefficients $d_{\alpha\beta}$ are the degrees of the map $$ S^{n-1}\to X_{n-1}\to S^{n-1} $$ where the first map is the attaching map of $e^n_\alpha$ to the $n-1$-skeleton, and the second map is collapsing the $X_{n-1}$ except for $e^{n-1}_\beta$.

What if you had a CW-complex consisting of a $0$-cell, with two $1$-cells attaching to it to make a figure-eight, and then attached a $2$-cell to one of the one cells to get a figure-eight with a filled in loop.

If I wanted to compute the coefficient in $d_2(e^2)$ of $e^1_\gamma$ where $e^1_\gamma$ is the $1$-cell I filled in, If I follow the composition above, the second map would collapse the unfilled loop to a point, but then it seems like I end up with $D^2$, not $S^1$. Am I doing something wrong?

$\endgroup$

1 Answer 1

1
$\begingroup$

You're left with $S^1$. You've already mapped the boundary of the $2$-cell to itself, and the other $1$-cell is collapsed to the $0$-cell.

$\endgroup$
3
  • $\begingroup$ Thanks Ted. After attaching to the $2$-cell to one of the loops, am I wrong in picturing a figure-eight with one loop filled in, like $D^2\vee S^1$? Wouldn't collapsing the unfilled $1$-cell to the $0$-cell give $D^2$? $\endgroup$ Sep 10, 2014 at 2:52
  • $\begingroup$ No, that loop just disappears. You're mapping $D^2$ to its boundary. You're thinking, I guess, of collapsing the loop and forgetting to do that in the $1$-skeleton. $\endgroup$ Sep 10, 2014 at 3:13
  • $\begingroup$ Oh shoot, I was replacing $X_{n-1}$ in my head with $D^2\cup_f X_{n-1}$ for the attaching map $f$. $\endgroup$ Sep 10, 2014 at 3:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .