I know the theory behind cellular homology, but I'm having troubles in the actual computation. The cellular boundary formula states that: $$d(e^n_{\alpha})=\sum_{\beta}d_{\alpha \beta}e^{n-1}_{\beta}$$ Where, if $n>1$: $$d_{\alpha\beta}=\deg(\varphi_{\alpha\beta})$$ $$\varphi_{\alpha\beta}: S_{\alpha}^{n-1}\xrightarrow{\text{attaching map}}X^{n-1}\xrightarrow{\pi}\frac{X^{n-1}}{X^{n-1}-e^{n-1}_{\beta}}\cong S_{\beta}^{n-1}$$ Where the last homeomorphism is defined by $\Phi_{\beta}:D^{n-1}_{\beta}\to X$ (the characteristic map of $e^{n-1}_{\beta}$). In fact we can restrict $\Phi_{\beta}$ in the codomain to $X^{n-1}$ and then $\Phi_{\beta}$ passes to the quotients(as a homeomorphism!): $$\tilde\Phi_{\beta}:\frac{D^{n-1}_{\beta}}{S^{n-2}_{\beta}}\to\frac{X^{n-1}}{X^{n-1}-e^{n-1}_{\beta}} $$ So: $$\tilde\Phi_{\beta}^{-1}:\frac{X^{n-1}}{X^{n-1}-e^{n-1}_{\beta}}\to\frac{D^{n-1}_{\beta}}{S^{n-2}_{\beta}}\cong S^{n-1}_{\beta} $$
Where the last homeomorphism is canonical.
This is crystal clear! But when I read online examples of (non-trivial) computation, I just can't understand what's going on. For example, sometimes I see the attaching map represented as a "word", and I don't understand what this representation means. I didn't manage to find online any non-trivial examples of computation that actually shows in detail how to calculate these degrees rigorously. (Maybe I just lack some intuition about the degree)
Now I'll try to show how I would compute the homology the thorus.
I'll choose the standard CW-decomposition of the thorus, so I have one $0$-cell,two $1$-cells, one $2$-cell:
$$0\leftarrow \mathbb{Z}\xleftarrow{d_1} \mathbb{Z}^2 \xleftarrow{d_2} \mathbb{Z}\leftarrow 0$$
Since there is only one $0$-cell we have $d_1=0$. Now we compute $d_2$. Let $\color{red}{e^1_1}$,$\color{blue}{e^1_2}$ be the two $1$-cells
The coefficient $d_{11}$: $$d_{11}=\deg\left(\varphi_{11}:S_{1}^{1}\xrightarrow{\text{attaching map}}X^{1}\xrightarrow{\pi}\frac{X^{1}}{X^{1}-e^{1}_{\beta}}\cong S_{1}^{1}\right)$$
Intuitively the attaching map attachs the $2$-disk along the $1$-cells(intuitively we can divide the circumference in $4$ quarters; two opposite quarters are sent in the same $1$-cell and the four junction points of the quarters are sent in the only $0$-cell ). Now since the canonical projection collapse the blue $1$-cell and the $0$-cell in one point, the map $\varphi_{11}:S^1_1\to S^1_1$ is:
Now:
$$\deg(\varphi_{11})=H_1(\varphi_{11})[\mathcal{C}]=[\varphi_{11}\circ \mathcal{C}]$$
Where $\mathcal{C}(x_0,x_1)=e^{2\pi ix_0}$ is the canonical generator of $H_1(S^1)$. Now I don't know how to proceed. According to the results $\deg(\varphi_{11})$ should be $0$, so $\varphi_{11}\circ \mathcal{C}$ should be a boundary. But I don't know if this is the right way.
Thank you in advance!