Use the transfer sequence for the covering \(S^\infty \to \mathbb{RP}^\infty\) to compute \(H_n(\mathbb{RP}^\infty; \mathbb{Z}_2)\).
Solution
Let \[ p:S^\infty\longrightarrow\mathbb{RP}^\infty \] be the antipodal double cover, and use \(\mathbb Z_2\) coefficients throughout. Hatcher’s transfer sequence is \[ \cdots\to H_n(\mathbb{RP}^\infty) \overset{\tau_*}{\longrightarrow}H_n(S^\infty) \overset{p_*}{\longrightarrow}H_n(\mathbb{RP}^\infty) \overset{\delta}{\longrightarrow}H_{n-1}(\mathbb{RP}^\infty)\to\cdots . \]
1. Since \(S^\infty\) is contractible,\[ H_n(S^\infty;\mathbb Z_2)=0\quad(n>0), \qquad H_0(S^\infty;\mathbb Z_2)=\mathbb Z_2. \]
The infinite sphere \(S^\infty\) is contractible, so it has the homology of a point.
2. For every \(n\ge2\), the connecting map is an isomorphism\[ \delta:H_n(\mathbb{RP}^\infty;\mathbb Z_2) \xrightarrow{\cong} H_{n-1}(\mathbb{RP}^\infty;\mathbb Z_2). \]
The portion of the transfer sequence around these groups is \[ 0\longrightarrow H_n(\mathbb{RP}^\infty) \overset{\delta}{\longrightarrow}H_{n-1}(\mathbb{RP}^\infty) \longrightarrow0, \] since the adjacent positive-dimensional homology groups of \(S^\infty\) vanish.
3. The same conclusion holds for \(n=1\):\[ H_1(\mathbb{RP}^\infty;\mathbb Z_2)\cong H_0(\mathbb{RP}^\infty;\mathbb Z_2)\cong\mathbb Z_2. \]
At the bottom of the transfer sequence we have \[ 0\to H_1(\mathbb{RP}^\infty) \overset{\delta}{\longrightarrow}H_0(\mathbb{RP}^\infty) \overset{\tau_*}{\longrightarrow}H_0(S^\infty) \overset{p_*}{\longrightarrow}H_0(\mathbb{RP}^\infty)\to0. \] Both spaces are connected, so \(p_*\) is the identity isomorphism on \(H_0\cong\mathbb Z_2\). Exactness therefore forces \(\tau_*=0\) and \(\delta\) to be an isomorphism.
4. Therefore\[ \boxed{H_n(\mathbb{RP}^\infty;\mathbb Z_2)\cong\mathbb Z_2\quad\text{for every }n\ge0.} \]
The degree-zero group is \(\mathbb Z_2\) because \(\mathbb{RP}^\infty\) is connected. Apply the isomorphisms of <1>2–<1>3 inductively.