Let \(X\) be a topological space obtained by attaching a 2-cell to \(\RP^2\) via some map \(f: S^1 \to \RP^2\) .
What are the possibilities for the homology \(H_* (X; Z)\)?
Solution
1. Up to homotopy, there are exactly two possible attaching maps\[ f:S^1\to\RP^2: \]the nullhomotopic map and a map representing the nonzero element of\[ \pi_1(\RP^2)\cong\ZZ/2\ZZ. \]
The homotopy classes of based loops in \(\RP^2\) form \(\pi_1(\RP^2)\cong\ZZ/2\ZZ\). Since the target is path-connected, free homotopy classes of maps \(S^1\to\RP^2\) are conjugacy classes in this group. The group is abelian, so its two elements give exactly two free homotopy classes.
Attaching a cell along homotopic attaching maps produces homotopy-equivalent adjunction spaces. Therefore the homology of \(X\) depends only on which of these two classes contains \(f\).
2. Use the standard CW structure on \(\RP^2\) with one cell in each dimension \(0,1,2\).Its cellular chain complex is\[ 0\longrightarrow\ZZ \xrightarrow{\;2\;}\ZZ \xrightarrow{\;0\;}\ZZ \longrightarrow0. \]
The \(1\)-skeleton is \(S^1\). The original \(2\)-cell of \(\RP^2\) is attached by the degree-\(2\) map \(S^1\to S^1\), so its cellular boundary is multiplication by \(2\). There is one \(0\)-cell, hence the boundary \(C_1\to C_0\) is zero.
3. If \(f\) is nullhomotopic, then the cellular chain complex of \(X\) in positive degrees is\[ 0\longrightarrow\ZZ^2 \xrightarrow{\;(2\;0)\;}\ZZ \xrightarrow{\;0\;}\ZZ \longrightarrow0. \]
2. With respect to the basis consisting of the original and new \(2\)-cells,\[ d_2(u,v)=2u. \]
4. In the nullhomotopic case,\[ H_k(X;\ZZ) \cong \begin{cases} \ZZ,&k=0,\\ \ZZ/2\ZZ,&k=1,\\ \ZZ,&k=2,\\ 0,&k\ge3. \end{cases} \]
For the differential in 3, \[ \ker d_2=\{(0,v):v\in\ZZ\}\cong\ZZ, \] so \(H_2(X;\ZZ)\cong\ZZ\). Also \[ H_1(X;\ZZ) = \ZZ/\operatorname{im}(d_2) = \ZZ/2\ZZ. \] The space is connected, so \(H_0\cong\ZZ\), and it has no cells above dimension \(2\). Equivalently, in this case \(X\simeq\RP^2\vee S^2\).
5. If \(f\) represents the nonzero element of \(\pi_1(\RP^2)\), then the cellular chain complex of \(X\) in positive degrees is\[ 0\longrightarrow\ZZ^2 \xrightarrow{\;(2\;1)\;}\ZZ \xrightarrow{\;0\;}\ZZ \longrightarrow0. \]
By cellular approximation, take the nontrivial attaching map to land in the \(1\)-skeleton \(S^1\). The generator of \(\pi_1(S^1)\cong\ZZ\) maps to the nonzero class modulo \(2\) in \[ \pi_1(\RP^2)\cong\ZZ/2\ZZ, \] so we may choose the degree-\(1\) loop on the \(1\)-skeleton as a representative. Hence the new \(2\)-cell contributes cellular boundary \(1\), while the original \(2\)-cell contributes \(2\). Thus \[ d_2(u,v)=2u+v. \]
6. In the nontrivial case,\[ H_k(X;\ZZ) \cong \begin{cases} \ZZ,&k=0,2,\\ 0,&k\ne0,2. \end{cases} \]
The map \[ d_2:\ZZ^2\to\ZZ, \qquad (u,v)\longmapsto2u+v \] is surjective, so \(H_1(X;\ZZ)=0\). Its kernel is \[ \{(u,-2u):u\in\ZZ\}\cong\ZZ, \] so \(H_2(X;\ZZ)\cong\ZZ\). Again \(H_0\cong\ZZ\) and all groups above degree \(2\) vanish.
7. These are the only two possibilities for \(H_*(X;\ZZ)\).
By 1 every attaching map belongs to exactly one of the two homotopy classes, and 4 and 6 compute the homology in those two cases.