Induced map between zeroth homology groups an isomorphism?

Question

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.

Answer 1

A map $\Delta^0 \to X$ is just a point $x\in X$, so we can identify $C_0(X) = \mathbb ZX$ (free $\mathbb Z$ module on basis $X$). What does it mean for two points $x, x'\in C_0(X)$ to become identified in $H_0(X)$? It means that there is a map $p: I \cong \Delta^1 \to X$ such that $p(0) - p(1) = x - x'$, and so $p(0) = x$ and $p(1) = x'$. That is, $x$ and $x'$ are in the same path component of $X$. So, $H_0(X) = \mathbb Z \pi_0(X)$ where $\pi_0(X)$ is the set of path components of $X$.

Thus, if any map $X\to Y$ induces a $\pi_0$ isomorphism, it follows that it also induces an $H_0$ isomorphism. If both spaces are path connected, this is automatic.

All of this goes through with any coefficient ring $R$ in place of $\mathbb Z$, by the way.