Here we examine

Here we examine

\[ H_1\RP^2 \mapsvia{\delta_1} H_0 \del M \mapsvia{(i^0, j^0)} H_0 M \oplus H_0 D^2 \mapsvia{l^0 - r^0} H_0\RP^2 \mapsvia{\delta_0} 0\\ \ZZ_2 \mapsvia{\delta_1} \ZZ \mapsvia{(i^0, j^0)} \ZZ \oplus \ZZ \mapsvia{l^0 + r^0} H_0\RP^2 \mapsvia{\delta_0} 0 \]

Since there is no nontrivial homomorphism from \(\ZZ_2 \to \ZZ\), we have \(\delta_1 = 0\).

We also have \(\delta_0 = 0\) and \(\ker \delta_0 = H_0 \RP^2 = \im l^0 + r^0\) making \(l^0 + r^0\) surjective, so by the first isomorphism theorem we have \(H_0 \RP^2 \cong \frac{\ZZ \oplus \ZZ}{\ker l^0 + r^0} = \frac{\ZZ \oplus \ZZ}{\im (i^0, j^0)}\)

By a similar argument used earlier, the double covering of the boundary circle \(\del M\) over \(S^1\) yields the map \((i^0, j^0): \ZZ \into \ZZ \oplus \ZZ\) given by \(x \mapsto (2x, 2x)\) with

Summary:

With all of this information, we finally have

And so we find \(H_*(\RP^2) = \ZZ \delta_0 + \ZZ_2\delta_1\)