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For the graph map $(1,f):M\to M\times M$, the cohomology class of the diagonal satisfiesIf $f$ is fixed-point-free, its graph misses the diagonal, so the pullback vanishes. This proves the Lefschetz fixed-point theorem for closed oriented manifolds.

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

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.

The Lefschetz fixed-point theorem says that a continuous self-map $f$ of a compact triangulable space has a fixed point whenever its Lefschetz numberis nonzero. In the smooth nondegenerate case, the Lefschetz-Hopf fixed-point theorem expresses this number as the sum of the local fixed-point indices.

Identify $H_1({\mathbb{R}}^2/{\mathbb{Z}}^2;\mathbb{Z})$ with ${\mathbb{Z}}^2$ by sending $v\in {\mathbb{Z}}^2$ to the class of the loop $t\mapsto tv\mathrm{mod}{\mathbb{Z}}^2$. The image of this loop under $f$ lifts from $0$ to the path $t\mapsto \tilde{f}(tv)$, whose endpoint is $\tilde{f}(v)=A(v)$. Its homology class is therefore $A(v)$. Thus $A=f_{\ast }$ under this identification, and in particular $A$ is a homomorphism represented by an integer matrix.

The expansion inequality implies, after taking a derivative, thatfor every tangent vector. The supplied linear-algebra fact shows that every eigenvalue of $D{\tilde{f}}_x$ has modulus at least $\mu >1$, so $1$ is not an eigenvalue. Every fixed point of $f$ is consequently nondegenerate and contributes local index $+1$ or $-1$.

On the torus, the induced maps on $H_0,H_1,H_2$ have traces $1,\mathrm{tr}A,\det A$. HenceThe original expansion inequality applied to lattice vectors gives $|Av|\ge \mu |v|$ for $v\in {\mathbb{Z}}^2$, and homogeneity and density extend it to all $v\in {\mathbb{R}}^2$. If ${\lambda }_1,{\lambda }_2$ are the possibly complex eigenvalues of $A$, then $|{\lambda }_j|\ge \mu$, soSince every fixed point contributes an index of absolute value one, the Lefschetz-Hopf formula gives

Its homology and chain formulas areFor Lefschetz fixed-point theorem, argue contrapositively: after barycentric subdivision, a fixed-point-free map has a simplicial approximation such that no simplex meets its image. Every diagonal coefficient on chains is then zero, so $L(f)=0$.

The antipodal map on $S^n$ has degree $(-1{)}^{n+1}$. For even $n$ this is $-1$, whereas the identity has degree $1$, so they are not homotopic.

If the tangent field $v$ were nowhere zero, thenwould stay on the sphere because $x\cdot v(x)=0$, and would homotope the identity to the antipodal map. This is impossible for even $n$, so $v$ has a zero.