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I am struggling to understand the following example in Hatcher's $\textit{Algebraic Topology}$. In section 2.2, it is:

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where we have some map definitions:

enter image description here

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!

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What Hatcher means by "$2$-cell attached by the product of commutators $[a_1,b_1]\cdots [a_g,b_g]$" is that the attaching map of the $2$-cell, i.e. the map $\varphi:S^1 \to M_g^1$ says that the boundary of the $2$-cell is identified with the loop $[a_1,b_1]\cdots [a_g,b_g]$ in the $1$-skeleton. I think you are confusing the more general picture of attaching maps with this situation where Hatcher is giving an explicit recipe for constructing $M_g$. The intuition is that if I take a flat piece of rubber as my $2$-cell I will attach it to my $1$-skeleton by molding its boundary into a $2g$-gon, order the sides $1,2,3,...,2g$ consecutively, then attach 1 to $a_1$, 2 to $b_1$, 3 to $a_1^{-1}$, 4 to $b_1^{-1}$, etc. in that fashion following the word $[a_1,b_1]\cdots [a_g,b_g]$.

The resolution of your second confusion comes from the fact that here we are considering homology and not homotopy groups. Remember that for homology we work with the cellular chain groups $C_i(M_g)$ in the complex $$ 0 \longrightarrow C_2(M_g) \overset{d_2}{\longrightarrow} C_1(M_g) \overset{d_1}{\longrightarrow} C_0(M_g) \longrightarrow 0 $$ where $C_i(M_g)$ is the free Abelian group generated by the $i$-cells of $M_g$. This chain complex agrees with the one shown in Example 2.36 since there is only one $2$-cell, $2g$ $1$-cells, and one $0$-cell. The map $d_2$ is the boundary operator and this is defined in Hatcher for cellular homology in such a way that $d_2$ of the $2$-cell is given by $$a_1 + b_1 - a_1 - b_1 + a_2 + b_2 - a_2 - b_2 +\ldots a_g+b_g - a_g - b_g = 0$$ where the signs depended on how the characteristic map of the $2$-cell attached to the $1$-skeleton. In particular, the signs come from the degrees of the maps $\Delta_{\alpha \beta}$. Since $d_2$ is zero on the generator of $C_2(M_g)$ then it is identically the zero homomorphism.

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    $\begingroup$ Does this answer whether the (delta) maps are homotopic to constants? $\endgroup$
    – Gil
    Jul 29 '16 at 10:03

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