Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 11F / Solution

An abstract smooth surface is a Hausdorff second-countable topological space with an atlas of charts to open subsets of ${\mathbb{R}}^2$ whose transition maps are smooth. It is orientable when it has such an atlas for which every transition map has positive Jacobian determinant. A map $f:S_1\to S_2$ is a smooth map when every coordinate representationis smooth wherever it is defined.

The map $a:C\to C$ is a smooth involution with no fixed point: the equationwould force $x=y=0$, which is impossible on $x^2+y^2=1$. For each $p\in C$, choose a sufficiently small coordinate neighbourhood $U_p$ such thatThen the quotient projection restricts to a homeomorphismTransporting a smooth chart ${\varphi }_p:U_p\to {\mathbb{R}}^2$ across this homeomorphism gives the quotient chartOn an overlap, a lift lies either in another chosen neighbourhood or in its image under $a$. The corresponding transition map is therefore a transition map on $C$, possibly composed with the diffeomorphism $a$, and is smooth. Since this is a free action of the finite group $\{1,a\}$, the quotient is Hausdorff and second countable. This constructs the smooth quotient by a free finite group action, and in these charts $\pi$ is locally the identity. Hence $\pi$ is a local diffeomorphism, in particular smooth.

To test orientability, parametrize the cylinder byThe involution acts in these coordinates aswhose derivative has determinant $-1$. Thus $a$ is an orientation-reversing diffeomorphism of the cylinder.

If the quotient surface $S$ were orientable, its orientation would pull back through the local diffeomorphism $\pi$ to an orientation of $C$. The identity $\pi \circ a=\pi$ would then force $a$ to preserve that pulled-back orientation, contradicting the negative determinant above. Therefore

Solved by gpt-5.6-sol high.