Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 / 4

Let $m=\dim M$. Choose a homogeneous basis $e_{q,i}$ of $H^q(M;\mathbb{Q})$ and its Poincare dual basis $e_{q,i}^{\ast }\in H^{m-q}(M;\mathbb{Q})$, normalized byWith the product orientation, the cohomology class of the diagonal isIndeed, multiplying this class by $\alpha \times \gamma$ and evaluating on $[M\times M]$ gives $⟨\alpha ⌣\gamma ,[M]⟩$, which characterizes the Poincare dual of $\Delta$.

Pulling back along the graph map $(1,f):M\to M\times M$ and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph is disjoint from $\Delta$. Represent ${\epsilon }_{\Delta }$ with support in a tubular neighbourhood disjoint from the graph; its pullback is zero, so $L(f)=0$. Contrapositively, $L(f)\ne 0$ implies that $f$ has a fixed point, the Lefschetz fixed-point theorem.

Now let $M$ be three disjoint circles. A homeomorphism permutes their three components. Its action on $H^1(M;\mathbb{Z})\cong {\mathbb{Z}}^3$ is a signed permutation matrix. A component fixed setwise contributes to the trace by the degree of the corresponding circle homeomorphism. An orientation-reversing circle homeomorphism has a fixed point, so fixed-point-freeness forces that degree to be $+1$. Nonfixed components contribute zero. The trace is therefore the number of fixed points of a permutation of three objects, and

For a compact manifold $N$ with boundary, let $D(N)=N_{+}{\cup }_{\partial N}{N}_{-}$ be its double and define $F$ by applying $f$ on both copies. The Mayer–Vietoris sequence for this decomposition is natural under $F$. Alternating traces in a finite-dimensional exact sequence sum to zero, so the two copies of $N$ contribute twice and their intersection contributes with the opposite sign:This is the Lefschetz number of a doubled map.

Finally let $N$ be a pair of pants. If $f$ is fixed-point-free, so are its double $F$ and its boundary restriction. The Lefschetz fixed-point theorem and the displayed identity give $L(F)=L(f{|}_{\partial N})=0$, hence $L(f)=0$. Since $N$ is connected and $H^i(N;\mathbb{Q})$ is nonzero only for $i=0,1$,Suppose the boundary permutation $\sigma$ had a fixed component. The restriction there is a fixed-point-free circle homeomorphism and thus has degree $+1$, forcing $f$ to preserve the surface orientation. The homology action of a pair-of-pants homeomorphism would then have trace $\#\mathrm{Fix}(\sigma )-1$, which is $2$ for the identity permutation and $0$ for a transposition. Neither is $1$. Therefore $\sigma$ has no fixed component; a permutation of three objects with no fixed point is a three-cycle. Thus