# Cellular boundary map for real projective space

I am trying to compute the cellular boundary map for real projective space: $\mathbb{R}P^n$. But, I am stuck. Any help will be appreciated. Here's what I have got so far.

I want the map from $d: H_n(X^n,X^{n-1}) \to H_{n-1}(X^{n-1},X^{n-2})$. Now, both the $H$'s are $\mathbb Z$ due to the cell structure. To calculate the effect of $d$ on the $e^n$ that I attach to $X^{n-1}$, I have to compute the degree of the following map: $$S^{n-1} \to \mathbb{R}P^{n-1} \to \mathbb{R}P^{n-1}/\sim \to S^{n-1}$$

Now, $\mathbb{R}P^{n-1}/\sim$ is got by collapsing $X^{n-1}$ except the (only) $n-1$-cell, so this is just $S^{n-1}$. Also, $\mathbb{R}P^{n-1}$ is $S^{n-1}$ with antipodal maps collapsed. So the first map can go like this: first we collapse the equator of $S^{n-1}$ to get a wedge of spheres and then pass to the quotient. So the map ultimately looks like this $x \mapsto$ the equivalence class $\lbrace (x,0), (0,-x) \rbrace$ in $\mathbb{R}P^{n-1}$.

And, now I am stuck. I have no idea how to go further. Thank you. [Also the one question similar to mine has not been answered.]

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Do you know about local degree? You've got a two to one mapping. One of the maps is the antipodal map, which has degree $(-1)^n$. And the other is (homotopic to) the identity, which has degree $1$. The degree of the map is then $1+(-1)^n$.