Action of a map on homology I'm currently studying algebraic topology from Hatcher's text, and I came across the following problem from an old qualifying exam:

The coefficient sequence $0 \rightarrow \mathbb{Z} \xrightarrow{p} \mathbb{Z} \xrightarrow{r} \mathbb{Z}/p \rightarrow 0$ induces a long exact sequence with boundary map $\overline{\beta}_p:H_k(X;\mathbb{Z}/p) \rightarrow H_{k-1}(X;\mathbb{Z})$. We define $\beta_p = r_{\ast}\overline{\beta}_p:H_k(X;\mathbb{Z}/p) \rightarrow H_{k-1}(X;\mathbb{Z}/p)$. Compute the action of $\beta_2$ on $H_{\ast}(\mathbb{R}P^{\infty};\mathbb{Z}/2)$.

I'm not quite sure how to approach this problem; namely, from my reading of Hatcher's text, I'm not familiar with what's meant by computing the action of a map on the homology groups. Hatcher discusses related ideas such as the action of $\pi_1$ on higher homotopy groups, but how can I compute the action specified in this problem?
Any thoughts would be appreciated.
Thanks!
 A: Consider the exact sequence
\begin{align}\ldots\longrightarrow H_{k}(\Bbb RP^\infty;\Bbb Z)\longrightarrow H_{k}(\Bbb RP^\infty;\Bbb Z/2)\overset{\bar{\beta_2}}{\longrightarrow}H_{k-1}(\Bbb RP^\infty;\Bbb Z)\overset{f}{\longrightarrow}H_{k-1}(\Bbb RP^\infty;\Bbb Z)\\ \overset{r_{\ast}}{\longrightarrow} H_{k-1}(\Bbb RP^\infty;\Bbb Z/2)\longrightarrow \ldots\end{align}
Suppose $k\ge 2$ first.

*

*If $k$ is even, then $H_{k}(\Bbb RP^\infty;\Bbb Z)\cong 0$ so that $\bar{\beta_2}$ is injective and hence an isomorphism because $H_{k}(\Bbb RP^\infty;\Bbb Z/2)$ and $H_{k-1}(\Bbb RP^\infty;\Bbb Z)$ are both isomorphic to $\Bbb Z/2$, which forces $f$ to be trivial so that $r_*$ is also an isomorphism. Therefore, $\beta_2=r_*\bar{\beta_2}$ is also an isomorphism.


*If $k$ is odd, then $H_{k-1}(\Bbb RP^\infty;\Bbb Z)\cong 0$, which means that $\bar{\beta_2}$ trivial so that $\beta_2=r_*\bar{\beta_2}$ must also be trivial.
Suppose $k=1$, then we are computing the map $\beta_2:H_1(\Bbb RP^\infty;\Bbb Z/2)\to H_0(\Bbb RP^\infty;\Bbb Z/2)$. The long exact sequence for this part looks like the following.
\begin{align}\ldots\longrightarrow H_1(\Bbb RP^\infty;\Bbb Z/2)\overset{\bar{\beta_2}}{\longrightarrow}H_0(\Bbb RP^\infty;\Bbb Z)\overset{f}{\longrightarrow} H_0(\Bbb RP^\infty; \Bbb Z)\overset{r_*}{\longrightarrow}H_0(\Bbb RP^\infty;\Bbb Z/2)\longrightarrow 0\end{align}
Since we have $H_1(\Bbb RP^\infty;\Bbb Z/2)\cong\Bbb Z/2$ and $H_0(\Bbb RP^\infty;\Bbb Z)\cong \Bbb Z$, the map $\bar{\beta_2}$ has to be trivial, and $\beta_2=r_*\bar{\beta_2}$ must also be trivial.
In conclusion, $\beta_2:H_k(\Bbb RP^\infty;\Bbb Z/2)\to H_{k-1}(\Bbb RP^\infty;\Bbb Z/2)$ is an isomorphism for $k$ even and trivial for $k$ odd.
