= Solution
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 $(\mathbb R^r,\mathbb R^r\setminus0)$ to its positive generator. The <Thom isomorphism theorem> states that this class exists uniquely with those fiber restrictions and that
$$
\boxed{H^q(B;\mathbb Z)\xrightarrow{\ \cong\ }H^{q+r}(E,E\setminus B;\mathbb Z),\qquad
a\longmapsto\pi^*a\smile U.}
$$
Here $B$ 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 $\boxed{\varepsilon_Y\in H_c^r(M;\mathbb Z)}$. 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 $M$. Set
$$
L(f)=\sum_k(-1)^k\operatorname{tr}(f^*:H^k(M;\mathbb Q)\to H^k(M;\mathbb Q)).
$$
Use 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 $K'$ admitting a simplicial approximation $F:K'\to K$ uniformly within $\delta/4$ of $f$. The subdivision <chain map> $s:C_*(K)\to C_*(K')$ 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 $f_*$ on <homology>. The trace identity yields $L(f)=0$. Taking the contrapositive proves
$$
\boxed{L(f)\ne0\ \Longrightarrow\ f\text{ has a fixed point}.}
$$
This 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>: $L(f)=\langle(1,f)^*\varepsilon_\Delta,[M]\rangle$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same <cohomology> maps, so their <Lefschetz numbers> agree.
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