Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 13 / 4

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

Take the double cover of $S^2$ branched at $2g+2$ points, with deck involution $\iota$. One explicit construction cuts the sphere along $g+1$ disjoint arcs pairing the branch points and cross-glues two copies along the cuts. Its Euler characteristic is $2\chi (S^2)-(2g+2)=2-2g$, so it is an oriented closed surface of genus $g$. Locally at a branch point the projection is $z\mapsto z^2$ and the involution is $z\mapsto -z$; thus the involution is nonidentity and orientation-preserving, also when $g=0$.

The invariant rational cohomology of a finite surface action identifies with the cohomology of its quotient. For this action, triangulate the quotient with the branch points as vertices and lift the triangulation. Quotient cochains identify with invariant cochains, and averaging over the two deck transformations makes taking invariants exact over $\mathbb{Q}$. Hence the invariant part of $H^1(S;\mathbb{Q})$ is $H^1(S^2;\mathbb{Q})=0$. Since ${\iota }^2=1$, its eigenvalues are $\pm 1$ and it is therefore $-I$ on the $2g$-dimensional $H^1$. Its actions on $H^0$ and $H^2$ are both the identity. ThusHomotopy invariance and the result just proved show that every map homotopic to this involution has fixed points, for every genus.

Let $P$ be a sphere with $g+1$ disjoint open disks removed, and form its double $S$ by gluing two copies along the boundary. Then $\chi (S)=2\chi (P)=2(1-g)$, so $S$ is an oriented closed surface of genus $g$. The involution $\iota$ swapping the two halves fixes precisely the $g+1$ seam circles.

Use disjoint collars with coordinates $(s,t)\in \mathbb{R}/\mathbb{Z}\times (-\eta ,\eta )$, where $\iota (s,t)=(s,-t)$. Choose a smooth even bump function $\chi (t)$ supported inside the collar, with $\chi (0)=1$, and choose $0<ϵ<1$. Define an orientation-preserving diffeomorphism $R$ on each collar by $R(s,t)=(s+ϵ\chi (t),t)$ and as the identity outside the collars. Scaling $ϵ$ from zero gives an isotopy from the identity to $R$, so $F=R\circ \iota$ is homotopic to $\iota$.

A fixed point of $F$ in a collar would satisfy $-t=t$, hence $t=0$. Its other coordinate would then require $s+ϵ=s$ modulo one, impossible for the chosen $ϵ$. Outside the collars, $F=\iota$ exchanges the interiors of the halves and has no fixed points. Therefore $\iota$ has fixed circles but is homotopic to the fixed-point-free map $F$. This is the fixed-point-free perturbation of a doubled-surface reflection; the construction includes genus zero, where the double is a sphere.