# Cellular homology boundary maps of a closed orientable surface of genus g

When computing homology of a closed orientable surface of genus g we get the following chain complex in cellular homology:

$0 \rightarrow \mathbb{Z} \xrightarrow{d_2} \mathbb{Z}^{2g} \xrightarrow{d_1} \mathbb{Z} \rightarrow 0$

My notes say that both $d_1$ and $d_2$ here are trivial, but I don't see why. Could I get some help understanding this?

Your question is missing some information regarding what cell decomposition you are using. But I think it's safe to assume, from the form of the chain complex that you have written, that you are perhaps thinking of attaching the boundary of the 2-cell in the "standard" fashion to the closed curve $$a_1 b_1 a_1^{-1} b_1^{-1} … a_g b_g a_g^{-1} b_g^{-1}$$ Let me write $A_1$ for the 1-chain corresponding to $a_1$, etc. It follows that $d_2$ maps the generator to $$A_1 + B_1 - A_1 - B_1 + … + A_g + B_g - A_g - B_g = 0$$ Also, there's only a single 0-cell $v$ representing a $0$-chain $V$, and each end of each oriented $1$-cell maps to $v$. So $d_1$ maps each of the $2g$ generators to $$V-V=0$$
It's possible that my assumption about the "standard" attachment is not true, in which case the argument is a bit more complicated. You have to use the orientation to conclude that each edge occurs exactly twice in the attachment, once with positive exponent and once with negative exponent. But in that case once again the $+1$ and $-1$ for that edge cancel, and $d_2$ is still zero.