Show that \(\RP^2 \lor S^1\) is not homotopy equivalent to a compact surface (possibly with boundary).
Solution
1. For\[ X=\RP^2\vee S^1, \]one has\[ \chi(X)=0, \qquad H_1(X;\ZZ)\cong\ZZ/2\ZZ\oplus\ZZ, \qquad H_2(X;\ZZ)=0. \]
The wedge has the CW structure obtained by adjoining to the standard CW structure on \(\RP^2\) one additional \(1\)-cell at the common basepoint. Thus it has one \(0\)-cell, two \(1\)-cells, and one \(2\)-cell, so \[ \chi(X)=1-2+1=0. \]
Reduced homology takes wedges of connected CW complexes to direct sums, hence \[ \widetilde H_i(X;\ZZ) \cong \widetilde H_i(\RP^2;\ZZ) \oplus \widetilde H_i(S^1;\ZZ). \] Using \[ H_1(\RP^2;\ZZ)\cong\ZZ/2\ZZ, \qquad H_1(S^1;\ZZ)\cong\ZZ, \] and \(H_2(\RP^2;\ZZ)=H_2(S^1;\ZZ)=0\) gives the stated groups.
2. If a compact connected surface \(S\) were homotopy equivalent to \(X\), then \(S\) would be one of\[ T^2, \qquad S^1\times[0,1], \qquad K, \qquad M, \]where \(K\) is the Klein bottle and \(M\) is the Möbius band.
Euler characteristic is a homotopy invariant for finite CW complexes, so 1 would give \[ \chi(S)=0. \] By the classification of compact connected surfaces, \[ \chi(\Sigma_{g,b})=2-2g-b \] for an orientable surface and \[ \chi(N_{k,b})=2-k-b \] for a nonorientable surface. The nonnegative integer solutions of these equations with Euler characteristic zero are \[ (g,b)=(1,0),(0,2) \] and \[ (k,b)=(2,0),(1,1), \] respectively. These are exactly the torus, annulus, Klein bottle, and Möbius band.
3. The torus, annulus, and Möbius band are not homotopy equivalent to \(X\).
The torus has \[ H_2(T^2;\ZZ)\cong\ZZ, \] whereas 1 gives \(H_2(X;\ZZ)=0\). Thus \(T^2\not\simeq X\).
Both the annulus and the Möbius band deformation retract onto a circle, so each has \[ H_1(-;\ZZ)\cong\ZZ. \] This differs from \[ H_1(X;\ZZ)\cong\ZZ/2\ZZ\oplus\ZZ. \] Hence neither is homotopy equivalent to \(X\).
4. The fundamental group\[ \pi_1(X) \cong \ZZ/2\ZZ * \ZZ \]contains a nontrivial element of order \(2\).
By Seifert–van Kampen for a wedge, \[ \pi_1(X) \cong \pi_1(\RP^2)*\pi_1(S^1) \cong \ZZ/2\ZZ * \ZZ. \] The canonical map from either free factor into a free product is injective. Therefore the nonidentity element of the factor \(\ZZ/2\ZZ\) remains an element of order \(2\) in \(\pi_1(X)\).
5. The Klein-bottle group is torsion-free.
Use the standard presentation \[ \pi_1(K) \cong G=\left\langle a,b\ \middle|\ aba^{-1}=b^{-1}\right\rangle. \] The relation gives \[ ab=b^{-1}a, \] so every element of \(G\) can be written in the form \[ b^m a^n \qquad(m,n\in\ZZ). \] Moreover \(G\) is the semidirect product \[ \ZZ\rtimes_{-1}\ZZ, \] with multiplication \[ (m,n)(m',n') = \bigl(m+(-1)^n m',\,n+n'\bigr), \] under the correspondence \((m,n)\leftrightarrow b^m a^n\). Thus this normal form is unique.
Suppose \(g=b^m a^n\) has finite order \(r>0\). Projection onto the second coordinate defines a homomorphism \[ G\to\ZZ, \qquad b^m a^n\longmapsto n. \] From \(g^r=1\) we obtain \[ r n=0, \] so \(n=0\). Hence \(g=b^m\). But \(\langle b\rangle\cong\ZZ\), so \(b^m\) has finite order only when \(m=0\). Therefore \(g=1\), proving that \(G\) is torsion-free.
6. The Klein bottle is not homotopy equivalent to \(X\).
A homotopy equivalence induces an isomorphism on fundamental groups. By 4, \(\pi_1(X)\) contains a nontrivial element of order \(2\), while by 5, \(\pi_1(K)\) is torsion-free. Thus \[ \pi_1(X)\not\cong\pi_1(K), \] so \(X\not\simeq K\).
7. Therefore \(\RP^2\vee S^1\) is not homotopy equivalent to any compact surface, with or without boundary.
By 2, a compact connected surface homotopy equivalent to \(X\) would have to be one of four surfaces. Steps 3 and 6 exclude all four.