# Calculating cellular homology, how to relate degree of boundary map with hand waving arguement

I am confused about where the degree of the boundary mapping comes into play, the examples I have seen when calculating cellular homology seem to do instead use some (what seems to be) hand waving to give the boundary maps. Is the following correct and if so could you explain how it related to the degree of the boundary mapping?

The example I am working through at the moment is $$(S^2\coprod S^2)/\sim$$ where $$n_1\sim n_2$$ and $$s_1 \sim s_2$$, i.e. we glue a sphere to another sphere at their north and south poles. Call this space $$X$$.

To calculate the cellular homology we first give this a CW structure. The easiest one I can think of consists of: two 0-cells the north and south pole which we call $$n$$ and $$s$$, two 1-cells each joining $$n$$ to $$s$$ which we will call $$a$$ and $$b$$, and two 2-cells $$U$$ and $$V$$ where $$U$$ is attached by glueing along $$a$$ then along $$-a$$.

This is the part I am uncomfortable with I am not sure how the following argument based on where the attaching map starts and finishes related to the degree of the boundary map.

Since we have two cells in each non trivial dimension, this gives use the following sequence of chains $$C_0(X)\leftarrow C_1(X)\leftarrow C_2(X)\leftarrow C_3(X)\leftarrow$$ as $$\mathbb{Z}^2\leftarrow\mathbb{Z}^2\leftarrow\mathbb{Z}^2\leftarrow -0$$.

Looking at the boundary maps, the map $$\partial_1:C_1(X)\rightarrow C_0(X)$$ sends $$a$$ to $$n-s$$ and $$b$$ to $$n-s$$. Therefore $$Im(\partial_1)\cong \mathbb{Z}$$ and hence $$ker(\partial_1)\cong \mathbb{Z}$$.

The map $$\partial_2:C_2(X)\rightarrow C_1(X)$$ sends $$U$$ to $$a-a=0$$ and $$V$$ to $$b-b=0$$, therefore $$\partial_2$$ is the zero map. Trivially $$\partial_3$$ is the zero map.

Now since we have the kernel and image of each of these maps we can easily calculate the homology, but again I don't see where we used the degree of the boundary map and my argument feels hand wavy.

• Your second paragraph seems to consist of an incomplete sentence, with a clause "... where $n_1$ and $n_2$ and $s_1$ and $s_2$..." that comes to an abrupt and unnatural halt. Commented Jan 10, 2020 at 15:41
• Thanks for pointing that out, there is meant to be a tilda in between them for an equivalence relation. I didn't realize typing ~ doesn't show up in mathjax. Is it clear now? Commented Jan 10, 2020 at 15:43
• Ah, that's better. And now it's also clearer that $n$ and $s$ denote north and south, which I did not previously understand. Commented Jan 10, 2020 at 16:24

Your calculation is correct, but to justify it a bit more using degrees you can proceed something like this. The idea is basically to use good notation by naming and using the characteristic map $$\chi_c$$ and attaching map $$f_c$$ of every cell $$c$$, and there is also a useful quotient map $$q_c$$.
Let's denote the skeleta as \begin{align*} X^{(0)} &= \{n,s\} \\ X^{(1)} &= X^{(0)} \cup a \cup b \\ X^{(2)} = X &= X^{(1)} \cup U \cup V \end{align*} The 1-cell $$a$$ comes equipped with a characteristic map $$\chi_a : D^1 \mapsto X^{(1)}$$. Similarly, the 2-cell $$U$$ comes equipped with a characteristic map $$\chi_U : D^2 \to X^{(2)}$$. Restricting $$\chi_U$$ to the boundary circle we obtain the attaching map $$f_U : S^1 \to X^{(1)}$$.
Consider also the quotient map $$q_a$$ defined on $$X^{(1)}$$ which collapses all of $$X^{(1)} - \text{interior}(a)$$ to a point. The quotient space is homeomorphic to a circle, and so we can write this homeomorphism as $$q_a : X^{(1)} \to S^1$$, chosen so that $$a$$ maps to the standard "counterclockwise" generator of $$H_1(S^1)$$.
Now simply compose $$S_1 \xrightarrow{f_U} X^{(1)} \xrightarrow{q_a} S^1$$ The degree of that composition $$q_a \circ f_U$$ is the coefficient you want, in this case equal to $$0$$.