To compute the homology of $\mathbb{RP}^n$ I need the following Lemma :
Lemma : $d_i(x_i) = \pm (\alpha_*(x_{i-1})+(-1)^i x_{i-1})$
Here $\alpha_*$ is the induced map the antipodal homeomorphism (which restricts to the skeletons $S^i \longrightarrow S^i$), $d_i$ is the composition map $j_{i-1} \circ \partial$, where $\partial$ is the boundary map of the pair and $j_{i-1}$ is the inclusion, i.e $H_i(S^i,S^{i-1}) \overset{\partial}{\longrightarrow} H_{i-1}(S^{i-1}) \overset{j_{i-1}}{\longrightarrow} H_{i-1}(S^{i-1},S^{i-2})$.
(In other words $d_i$ is the differential of the chain complex associated to the inductive decomposition of $S^{i}$ with $E_+^i, E_-^i$)
The proof flows except for the following. Once proven that $d_i \circ \alpha_* = (-1)d_i$, it proceeds like this "So $0 \ne \beta := \alpha_* (x_i) + (-1)^{i+1}(x_i) \in \text{Ker}(d_i)[...]$.
I do understand $\beta \in \text{Ker}(d_i)$ but I don't understand why $\beta \ne 0$. Any help or reference would be appreciated.
One can show that $d_ix_i\neq0$, (otherwise there is a problem in the dimension of the kernel of $di$).
A general element of the group is in the form $ax_i+b\alpha_*x_i$ and if it is in the kernel it satisfies $d_i(ax_i+b\alpha_*x_i)=0 \implies (a+(-1)^ib)d_ix_i=0 \implies a=(-1)^{(i+1)}b$
So a general element of the kernel is a multiple of $ax_i+b\alpha_*x_i$.