I am trying to calculate homology groups of the Klein bottle $K$ using cellular homology. $K$ has one $0$-cell, two $1$-cells and one $2$-cell:

$\require{AMScd}$ \begin{CD} x_0 @>{a}>> x_0 \\ @V{b}VV \circlearrowleft @A{b}AA \\ x_0 @>{a}>> x_0 \end{CD} (The circling arrow indicates an orientation on the 2-cell.)

So the cellular chain complex is of the form: \begin{equation} 0 \rightarrow \mathbb{Z} \xrightarrow[\text{}]{\delta_{2}} \mathbb{Z} \oplus \mathbb{Z}\xrightarrow[\text{}]{\delta_{1}} \mathbb{Z} \rightarrow 0\end{equation} , where $\delta_1$ and $\delta_2$ are the boundary maps. I have trouble with calculating those. In general, given an $i$-cell $\alpha$ with attaching map $\gamma_{\alpha}:S^{i-1} \rightarrow X^{(i-q)}$, and an $(i-1)$-cell $\beta$, we define $f_{\alpha,\beta}:S^{i-1} \rightarrow S^{i-1}$ to be the composition \begin{equation} S^{i-1} \xrightarrow[\text{}]{\gamma_{\alpha}} X^{(i-1)} \rightarrow X^{(i-1)}{/}X^{(i-2)}\xrightarrow[\text{}]{p_{\beta}} S_{\beta}^{i-1} \end{equation} I also know the Cellular Boundary Formula so I think that I need to calculate the degrees of $f_{a,x_0}$ and $f_{b,x_0}$ to get $\delta_1$ and to calculate the degrees of $f_{ab,a}$ and $f_{ab,b}$ to get $\delta_2$. Can someone explain how to calculate the degrees? Is there a general strategy to quickly do so?


1 Answer 1


For the 1-cell attaching maps the degree is easy: your attaching maps are constant because they are the maps $S^0 \to X^0 = \{x_0\}$ sending both endpoints of your $1$-cells to the $0$-cell. Thus both $\text{deg}f_{a,x_0}$ and $\text{deg}f_{b, x_0}$ are zero and thus $\delta_1 = 0$. It's worth noting that in general, if your $1$-skeleton ends up a wedge of circles the cellular boundary $\delta_1$ will always be zero, for this reason.

Now, I'll call the $2$-cell $e$, instead of $ab$ as you've done, so that when I refer to the $2$-cell it's a bit clearer (its attaching map will involve $a$'s and $b$'s so I don't want any confusion there).

$\delta_2$ also has a quick strategy for computation. The first step is usually to describe the attaching map for the $2$-cell in terms of the $1$-cells; in this case, we can say that the attaching map is $baba^{-1}$ (read off the edges in your polygon), corresponding to $\gamma: S^1 \to X^1 = S^1 \vee S^1$ that on the first quarter-circle of the domain traces the 1-cell $b$, on the second quarter-circle traces $a$, etc.

Now that we've done this, realize that $f_{e,a}$ is the attaching map restricted to $a$, so basically we're deleting $b$ from the formula for $\gamma$; this is the interpretation of the map $f_{\alpha,\beta}$ that you've described. This means that $\text{deg}f_{e,a}$ is the degree of the map described by $aa^{-1}$, which is constant. Similarly, $\text{deg}f_{e,b}$ is the degree of the map represented by $b^2$, which has degree $2$.

Therefore $\delta_1: \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ is the $0$ map and $\delta_2: \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$ sends $1 \mapsto (2,0)$.

  • $\begingroup$ thank you for such a nice answer. It makes sense now! $\endgroup$
    – billy192
    Commented Dec 18, 2019 at 8:15

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