# A pragmatic way of calculating with local degree

I understand the technical definition of local degree in a theoretical sense via isomorphism of $$H_n(U, U_i \backslash \{x_i\}) \cong H_n(S^n)$$ etc. However, I'm a bit confused by how exactly this correlates to how local degree is used in the calculation of cellular homologies. For example, I have been studying the example of $$\mathbb{RP}^2$$ which has the cell structure: $$e^0 \cup_{\phi} e^1 \cup_{\varphi} e^2$$.

I'd like to calculate the degree of the composition: $$S^{n-1} \xrightarrow{\varphi} X^{n-1} \rightarrow X^{n-1}/X^{n-2} \rightarrow \large \vee_j S^{n-1}\xrightarrow{q_j} S^{n-1}$$ Where $$q_j$$ is the map which sends all the $$S^{n-1}$$'s except the $$j^\text{th}$$ to the base point.

So in our example of $$\mathbb{RP^2}$$, to calculate the degree of the attaching map $$\varphi: \partial D^2 \cong S^1 \rightarrow X^1$$ we need to actually know how the space is constructed, not just its cell structure.

The map $$\varphi$$ will take the circle along the lone 1-cell $$e^1$$ twice, so to compute its degree via local degrees, I can choose a point in the one-cell* say $$y \in interior(e^1)$$ and then look at the discrete set $$\varphi^{-1}(y) \in S^1$$. Since we go along the same 1-cell twice, this will be a set of two points. $$\varphi^{-1}(y) = \{x_1, x_2\} \subset S^1$$

As I understand it, I then should look at how open neighbourhoods around these $$x_1, x_2$$ "look" compared to the open neighbour hood of $$y$$. I find that locally $$\varphi$$ will act just like a map sending a circle to a circle, in particular, it locally looks like either the identity map or the map which reverses the orientation of the circle. In this case, local to both $$x_1$$ and $$x_2$$ it will look like the identity which has degree 1.

Therefore, $$\deg{\varphi} = \deg(\varphi\vert_{U/{x_1}}) + \deg(\varphi\vert_{\tilde{U}/{x_2}}) = 1+1 = 2$$

My questions regarding this are the following:

1. To what extend does my choice of orientation matter, could I not have given $$S^1$$ the reverse orientation and end up with $$\deg(\varphi) = -2$$
2. If the 1-skeleton were to contain two 1-cells, but otherwise the construction was the same and I chose $$y$$ in this 1-cell which has nothing attached and $$\varphi^{-1}(y) = \phi$$ is the degree just 0?
3. If I'm attaching, say, a three cell or even a cell of higher dimension, it becomes less clear how I should envisage this orientation argument. How would one practically cope with this?