Suppose that $n,m$ are integers such that $n > m$. I know that the singular homology groups $H_0(\Bbb{R}P^n;\Bbb{Z})$ and $H_0(\Bbb{R}P^m;\Bbb{Z})$ are both isomorphic to $\Bbb{Z}$. Now suppose I look now at homology with $\Bbb{Z}/2\Bbb{Z}$ coefficients. My setup of maps is as follows: I have $g: S^n \to S^n$ and $g':S^m \to S^m$ antipodal maps and a continuous map $\phi : S^n \to S^m$ such that $g'\circ \phi = \phi \circ g$
Now from $\phi$, I obtain an induced map between quotient spaces $\psi : \Bbb{R}P^n \to \Bbb{R}P^m$. This $\psi$ in turn induces a map
$\psi_\ast : H_0(\Bbb{R}P^n;\Bbb{Z}/2\Bbb{Z})\longrightarrow H_0(\Bbb{R}P^m;\Bbb{Z}/2\Bbb{Z}).$
Why should $\psi_\ast$ be an isomorphism? I am trying to reason this out using the map $\begin{eqnarray*}f :& C_0(\Bbb{R}P^n;\Bbb{Z}/2\Bbb{Z}) &\longrightarrow C_0(\Bbb{R}P^m;\Bbb{Z}/2\Bbb{Z}) \\ &\sigma& \mapsto \psi \circ \sigma \end{eqnarray*} $
where $\sigma : \Delta^0 \to \Bbb{R}P^n$ is a singular $0$ - chain. Does $\psi_\ast$ being an isomorphism come from $f$ being one? I can see that $f$ is surjective (using lifting properties) while does injectivity of $f$ come from the fact that we are now talking of maps from a point to $\Bbb{R}P^n$?
Thanks.