= Solution
With the usual second-countability convention, an $n$-<manifold> is a <Hausdorff> second-countable space in which every point has a neighbourhood homeomorphic to an open subset of $\mathbb R^n$. A <local orientation of a manifold> at $x$ is a choice of generator of
$$
H_n(M,M\setminus\{x\};\mathbb Z)\cong\mathbb Z.
$$
Excision and a coordinate ball identify this relative <group> with $H_n(\mathbb R^n,\mathbb R^n\setminus\{0\})$. An <orientation of a topological manifold> is a coherent choice of such generators: near each point the choices are induced by one relative class on a neighbourhood, equivalently by an atlas with <orientation>-preserving transition maps.
For a <connected> closed oriented <manifold>, its <fundamental class> $[M]\in H_n(M;\mathbb Z)$ is the unique class mapping to every chosen local generator. Here closed means <compact> without boundary. <Poincare duality> states that cap product with this class is an isomorphism
$$
\boxed{-\frown[M]:H^k(M;\mathbb Z)\xrightarrow{\ \cong\ }H_{n-k}(M;\mathbb Z)}.
$$
Over rational coefficients it gives $b_k=b_{n-k}$. Therefore
$$
\chi(M)=\sum_k(-1)^kb_k=(-1)^n\sum_k(-1)^kb_k.
$$
When $n$ is odd this says $\chi=-\chi$, so \b[$\chi(M)=0$]. Torsion does not contribute to the Betti numbers or this <Euler characteristic>.
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