[proofplan] We prove the identity by reducing the evaluation of the Euler class to a signed zero count of a transverse section of the tangent bundle. The section transversality theorem permits us to choose a smooth vector field transverse to the zero section, so its zero set is finite and each zero has a well-defined local sign. The [zero-counting formula for the Euler class](/theorems/9775) identifies this signed count with $\langle e(TM),[M]\rangle$, while the Poincare-Hopf theorem identifies the same signed count with $\chi(M)$. [/proofplan] [step:Choose a transverse vector field and define its signed zero count] Let $\Gamma(TM)$ denote the space of smooth sections of the tangent bundle $TM\to M$. By the section transversality theorem, applied to the smooth vector bundle $TM\to M$ and the zero section submanifold $z(M)\subset TM$, there exists a smooth section $X\in \Gamma(TM)$ such that the corresponding smooth map \begin{align*} X:M\to TM \end{align*} is transverse to $z(M)$. The hypotheses of the section transversality theorem are satisfied because $M$ and $TM$ are smooth manifolds, $z(M)$ is an embedded smooth submanifold of $TM$, and smooth sections of $TM$ admit arbitrarily small smooth perturbations. Since $M$ is compact and $\operatorname{rank}(TM)=\dim M=n$, transversality implies that the zero set \begin{align*} Z(X):=\{p\in M:X(p)=0_p\} \end{align*} is a compact smooth submanifold of dimension $0$, hence a finite set. For each $p\in Z(X)$, let $\operatorname{ind}_p(X)\in \{1,-1\}$ denote the local orientation sign of the transverse intersection of $X(M)$ with $z(M)$ at $0_p\in TM$. Define the signed zero count \begin{align*} I(X):=\sum_{p\in Z(X)}\operatorname{ind}_p(X)\in \mathbb Z. \end{align*} [guided] We need a representative of the Euler class pairing that can be computed geometrically. The correct object is a transverse section of the oriented rank-$n$ bundle $TM\to M$. Let $\Gamma(TM)$ denote the space of smooth sections of $TM\to M$. A section of $TM$ is exactly a smooth vector field. By the section transversality theorem, applied to the smooth vector bundle $TM\to M$ and the embedded zero section submanifold $z(M)\subset TM$, there exists a smooth vector field $X\in \Gamma(TM)$ such that, as a map \begin{align*} X:M&\to TM, \end{align*} it is transverse to the zero section \begin{align*} z:M&\to TM. \end{align*} The theorem applies here because $M$ and $TM$ are smooth manifolds, $z(M)$ is an embedded smooth submanifold of $TM$, and smooth sections of a smooth vector bundle can be perturbed within the space of smooth sections. This transversality condition means that at every point $p\in M$ satisfying $X(p)=0_p$, the tangent spaces of the two submanifolds $X(M)$ and $z(M)$ span $T_{0_p}(TM)$. Define the zero set by \begin{align*} Z(X):=\{p\in M:X(p)=0_p\}. \end{align*} Since $X$ is transverse to $z(M)$, the preimage $X^{-1}(z(M))=Z(X)$ is a smooth submanifold of $M$ of dimension \begin{align*} \dim M-\operatorname{codim}_{TM}(z(M))=n-n=0. \end{align*} Because $M$ is closed, it is compact, and a compact smooth $0$-manifold is a finite set. Thus the following signed sum is finite. For each zero $p\in Z(X)$, the orientation of $M$ and the orientation of the tangent bundle $TM$ determine a local intersection sign \begin{align*} \operatorname{ind}_p(X)\in \{1,-1\}. \end{align*} This sign is the usual local index of the vector field at the nondegenerate zero $p$. We define the signed zero count by \begin{align*} I(X):=\sum_{p\in Z(X)}\operatorname{ind}_p(X). \end{align*} The rest of the proof identifies this one integer in two different ways: first with the Euler class pairing, and then with the Euler characteristic. [/guided] [/step] [step:Identify the Euler class pairing with the signed zero count] Apply the zero-counting formula for the Euler class, [citetheorem:9775], to the oriented smooth real vector bundle $TM\to M$ of rank $n$ and to the transverse section $X:M\to TM$. Its hypotheses are satisfied because $M$ is closed and oriented, $TM$ is oriented by the orientation of $M$, and $X$ is transverse to the zero section by construction. Therefore \begin{align*} \langle e(TM),[M]\rangle=\sum_{p\in Z(X)}\operatorname{ind}_p(X)=I(X). \end{align*} [/step] [step:Identify the same signed zero count with the Euler characteristic] We use the Poincare-Hopf theorem in the following form: if $Y\in\Gamma(TM)$ is a smooth vector field on a closed smooth manifold and all zeros of $Y$ are isolated and nondegenerate, then the sum of the local vector-field indices of $Y$ equals $\chi(M)$. When $M$ is oriented, the local vector-field index at a nondegenerate zero is the same sign as the transverse-intersection sign of the section $Y:M\to TM$ with the zero section. Applying this theorem to $Y:=X$, its hypotheses are satisfied because $M$ is closed and because transversality of $X$ to $z(M)$ makes every zero isolated and nondegenerate. Therefore \begin{align*} I(X)=\chi(M). \end{align*} [guided] The second identification of $I(X)$ uses the Poincare-Hopf theorem. The precise version needed is this: for a smooth vector field $Y\in\Gamma(TM)$ on a closed smooth manifold, if every zero of $Y$ is isolated and nondegenerate, then the sum of the local vector-field indices of $Y$ is the Euler characteristic $\chi(M)$. In the oriented case, the local index at a nondegenerate zero is computed by the same orientation sign as the transverse intersection of the section $Y:M\to TM$ with the zero section $z(M)\subset TM$. We apply this theorem with $Y:=X$. The manifold $M$ is closed by hypothesis. The zeros of $X$ are isolated because the previous step showed that \begin{align*} Z(X)=\{p\in M:X(p)=0_p\} \end{align*} is a finite $0$-dimensional submanifold. They are nondegenerate because transversality of $X$ to $z(M)$ says exactly that, in a local trivialization of $TM$ near a zero $p$, the local representative of $X$ has invertible derivative at $p$. Thus the local index appearing in Poincare-Hopf is precisely $\operatorname{ind}_p(X)$ as defined by the transverse-intersection orientation sign. Hence Poincare-Hopf gives \begin{align*} I(X)=\sum_{p\in Z(X)}\operatorname{ind}_p(X)=\chi(M). \end{align*} [/guided] [/step] [step:Conclude the evaluation formula] Combining the two identifications of $I(X)$ gives \begin{align*} \langle e(TM),[M]\rangle=I(X)=\chi(M). \end{align*} Hence the Euler class of the tangent bundle evaluates on the fundamental class of $M$ as the Euler characteristic of $M$. [/step]