# Simple cellular homology computation

Here's a very simple cellular homology computation that I'm a little confused about.

Put a CW structure on the closed disc $X=D^{2}$ with two zero-cells $v_{0},v_{1}$, two one-cells $e_{0},e_{1}$ connecting the zero cells, and one two-cell $f$ on the inside. The cellular chain complex would look like $$\cdots0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}^{2}\rightarrow\mathbb{Z}^{2}\rightarrow0,$$ where the first non-trivial map sends $f\mapsto e_{0}+e_{1}$ by the cellular boundary formula. The map $H_{1}(X^{1},X^{0})\rightarrow H_{0}(X^{0})$ should be the same as the usual boundary map, but how do we know whether $e_{0}\mapsto v_{0}-v_{1}$ or $e_{0}\mapsto v_{1}-v_{0}$? I suspect that it doesn't actually matter, because we're a free basis up to sign, but certainly if $e_{0}\mapsto v_{0}-v_{1}$, then we need $e_{1}\mapsto v_{1}-v_{0}$ to have a chain complex.

See here: the coefficients are the degree of the attaching map. Yes, your last sentence is correct: if $e_0 \mapsto v_0 -v_1$ then $e_1 \mapsto v_1 - v_0$.

• This is true for the 2-cells and higher, but we don't have a notion of degree on $S^{0}$, do we? My question is, how do we know whether to take $v_{0}-v_{1}$ or $v_{1}-v_{0}$?
– LCL
Mar 13, 2013 at 23:32

Your edges should be oriented, so that they have a "forward" direction. Then, the boundary map maps each edge to the "front" vertex minus the "back" vertex.

You also have to take this orientation into account when deciding the boundary map from the two cell. That is, if you switched the orientation of $e_1$, then the cellular boundary map from $H_2(X^2,X^1) \to H_1(X^1,X^0)$ would become $f \mapsto e_0 - e_1$.

• Huh, just looked at the date on this question. I wonder how this popped up in my feed.
– Carl
Sep 6, 2014 at 7:13