Map on homology induced by covering map $S^n \to \mathbb{R}P^n$ Given the standard covering map $p: S^n \to\mathbb{R}P^n $ and I want to find the map $p_{*}: H_n(S^n, R) \to H_n(\mathbb{R}P^n, R)$ (R is an arbitrary ring).
I introduce the following CW structures on the spaces. $\mathbb{R}P^n = e_n \cup e_{n-1} \ldots \cup e_0$ (one cell in each dimension) and $S^n = e_n^{(1)} \cup e_n^{(2)} \cup \ldots  $ (it's not that important what comes next, those $e_n$ are upper and lower halves of the sphere).
So I want to use cellular homology theory. Namely, I checked that $H_n(S^n, R) \cong R$ is generated  by the class $[\sigma_n^{(1)} + \sigma_n^{(2)}]$ where $\sigma_n^{(1),(2)}$ are the cell attaching maps for $e_n^{(1), (2)}$ and $H_n(\mathbb{R}P^n, R) \cong R$ (whenever it's nonzero) is generated by the class $[\tau_n]$  of attaching map of $e_n$.
Thus, $p_{*} [\sigma_n^{(1)} + \sigma_n^{(2)}] = p_{*} [\sigma_n^{(1)}]  + p_{*}[\sigma_n^{(2)}]  = \tau_n \pm \tau_n $. The problem is I can't figure out the sign here. So what I am asking is to help me with the sign.
 A: Let $R\sigma_n^{(1)}$ denote inclusion of the cell $e_n^{(1)}$ followed by multiplication by $-1$ on $S^n$.  Then $p_*(R\sigma_n^{(1)})=p_*(\sigma_n^{(1)})$, so the only thing we have to decide is which of $\sigma_n^{(2)}=\pm R\sigma_n^{(1)}$is true.
As $[\sigma_n^{(1)}+\sigma_n^{(2)}]$ is a homology class, we have $\partial(\sigma_n^{(1)})=-\partial(\sigma_n^{(2)})$.
So if $R\partial(\sigma_n^{(1)})=\partial(\sigma_n^{(1)})$, then $\partial(R\sigma_n^{(1)})=-\partial(\sigma_n^{(2)})$.  Then $\sigma_n^{(2)}=-R\sigma_n^{(1)}$ and $p_*(\sigma_n^{(2)})=-p_*(\sigma_n^{(1)})$. Then $$p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=0.$$
On the other hand, if $R\partial(\sigma_n^{(1)})=-\partial(\sigma_n^{(1)})$, then $\partial(R\sigma_n^{(1)})=\partial(\sigma_n^{(2)})$.  Then $\sigma_n^{(2)}=R\sigma_n^{(1)}$ and $p_*(\sigma_n^{(2)})=p_*(\sigma_n^{(1)})$. Then $$p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=2\tau_n.$$
Multiplication by $-1$ preserves orientation of the equator $S^{n-1}$ precisely when $n$ is even.  That is multiplication by $-1$ sends a generator of $H_{n-1}(S^{n-1})$ to $(-1)^{n}$ times itself.
To see this note that if we multiply just one pair of co-ordinates by $-1$, this is a $180^\circ$ rotation, hence homotopic to the identity.  Thus when $n$ is even, we have a sequence of homotopies from multiplication by $-1$ to the identity.  On the other hand if $n$ is odd, then we have a sequence of homotopies from multiplication by $-1$ to a reflection.
We conclude that $p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=0$ if $n$ is even and $p_*([\sigma_n^{(1)}+\sigma_n^{(2)}])=2\tau_n$ if $n$ odd.
