I am struggling to understand the following example in Hatcher's $\textit{Algebraic Topology}$. In section 2.2, it is:
where we have some map definitions:
and $d_n = j_{n−1}\partial_n$ for boundary operator $\partial_n: H_n(X^n,X^{n-1}) \longrightarrow H_{n-1}(X^{n-1})$ and $j_n$ is the quotient map $j_n:H_n(X^n) \longrightarrow H_n(X^n,X^{n-1})$. (For this example, $X = M_g$). Namely I don't understand the points:
- 2-cell attached by the product of commutators $[a_1,b_1]\cdots [a_g,b_g]$
- Also $d_2$ is 0 because each $a_i$ or $b_i$ appears with its inverse in $[a_1,b_1]\cdots[a_g,b_g]$ so the maps $\Delta_{\alpha\beta}$ are homotopic to constant maps.
In particular, my understanding of attaching a 2 cell is that you have an attachment map $f:\partial X^2 \rightarrow X_{1}$, so how does this relate to this product of commutators? Also, given that $a_i$ and $b_i$ along with their inverses appear in the product $$[a_1,b_1]\cdots[a_g,b_g] = a_1^{-1}b_1^{-1}a_1b_1 \cdots a_g^{-1}b_g^{-1}a_gb_g$$ why does this mean that $\Delta_{\alpha\beta}$ is homotopic to a constant map? And why does $\Delta_{\alpha\beta}$ being homotopic to a constant map mean that $d_2$ is 0?
Thanks so much!