Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 16 / 5

The surface group of a closed oriented genus-two surface $Y$ has presentationSend $a_1\mapsto (12)$, $a_2\mapsto (123)$ and $b_1,b_2\mapsto 1$. The relation holds and the images generate $S_3$, giving a surjective group homomorphism. Its normal kernel defines a connected six-sheeted regular covering space $\tilde{Y}\to Y$, with deck transformation group $S_3$.

The covering surface is closed and inherits an orientation. Multiplicativity of the Euler characteristic gives $\chi (\tilde{Y})=6\chi (Y)=6(-2)=-12$, so $2-2g=-12$ and $\boxed{g=7}$. A nonidentity deck transformation of a connected cover fixes no point: if it fixed one, uniqueness of path lifts would force it to be the identity everywhere. Thus this supplies the requested free $S_3$ action. It is a case of finite groups act freely on suitable closed orientable surfaces.

For an $n$-dimensional singular simplex $\sigma$ and a degree-$p$ singular cochain $\alpha$, use the front-face/back-face conventionand set it to zero for $p>n$. Extend linearly over $R$. Passing to homology classes of cycles and cohomology classes of cocycles defines the cap product $H_n(X;R)\times H^p(X;R)\to H_{n-p}(X;R)$.

For a continuous map $f$, the cochain pullback satisfies $(f^{\ast }\alpha )(\tau )=\alpha (f\circ \tau )$, and the induced chain map sends $\sigma$ to $f\circ \sigma$. Since composition commutes with restrictions to faces,The identity holds on chains, so it descends directly to classes and proves the naturality of the cap product: $f_{\ast }(x⌢f^{\ast }\alpha )=f_{\ast }(x)⌢\alpha$. No additional well-definedness verification is required here.