# Homology of the Klein bottle using cellular homology

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?

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 $$1 \mapsto (2,0)$$.