Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 21F (Algebraic Topology)

For a continuous self-map $f$ of a finite CW complex, its Lefschetz number over $\mathbb{Q}$ isThe Lefschetz fixed-point theorem states that $L(f)\ne 0$ implies that $f$ has a fixed point. Consequently every fixed-point-free self-map has Lefschetz number zero.

The space in the question is a cyclic point-gluing of two-spheres. Before gluing, the $n$ spheres have total Euler characteristic $2n$. Each of the $n$ point identifications lowers the Euler characteristic by one, soEquivalently, $X\simeq S^1\vee {\bigvee }_{j=1}^n{S}^2$, with rational homology dimensions $1,1,n$ in degrees zero, one, and two.

Suppose ${\varphi }^3={\mathrm{Id}}_X$. Since $\varphi$ has no fixed points, neither does ${\varphi }^2$: if ${\varphi }^2(x)=x$, then applying $\varphi$ gives $\varphi (x)={\varphi }^3(x)=x$. The Lefschetz theorem therefore gives

For each $q$, the averaging operatoris the projection of $H_q(X;\mathbb{Q})$ onto its $⟨\varphi ⟩$-invariant subspace, so its trace is an integer. Taking the alternating sum of these traces givesThe left side is an integer. By Euler-characteristic divisibility from a fixed-point-free cyclic action,

Now suppose ${\varphi }^2={\mathrm{Id}}_X$. The map is fixed-point-free, so $L(\varphi )=0$. On each rational homology group,projects onto the invariant subspace and has integral trace. Henceis an integer. Therefore