= Solution
Let $m=\dim M$. Choose a homogeneous basis $e_{q,i}$ of $H^q(M;\mathbb Q)$ and its <Poincare dual> basis $e_{q,i}^*\in H^{m-q}(M;\mathbb Q)$, normalized by
$$
\langle e_{q,i}\smile e_{q,j}^*,[M]\rangle=\delta_{ij}.
$$
With the product orientation, the <cohomology class of the diagonal> is
$$
\boxed{\varepsilon_\Delta=\sum_{q,i}(-1)^{mq}e_{q,i}^*\times e_{q,i}.}
$$
Indeed, multiplying this class by $\alpha\times\gamma$ and evaluating on $[M\times M]$ gives $\langle\alpha\smile\gamma,[M]\rangle$, 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>:
$$
\left\langle(1,f)^*\varepsilon_\Delta,[M]\right\rangle
=\sum_q(-1)^q\operatorname{tr}(f^*|H^q(M;\mathbb Q))
=L(f).
$$
If $f$ has no fixed point, its graph is disjoint from $\Delta$. Represent $\varepsilon_\Delta$ with support in a tubular neighbourhood disjoint from the graph; its pullback is zero, so $L(f)=0$. Contrapositively, \b[$L(f)\ne0$ 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
$$
\boxed{\operatorname{tr}(f^*|H^1(M;\mathbb Z))\in\{0,1,3\}.}
$$
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:
$$
\boxed{L(F)=2L(f)-L(f|_{\partial N}).}
$$
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 $i=0,1$,
$$
\operatorname{tr}(f^*|H^1(N;\mathbb Q))=1.
$$
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 $\#\operatorname{Fix}(\sigma)-1$, 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
$$
\boxed{f\text{ cyclically permutes the three boundary components}.}
$$
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