[proofplan] We pull back the diagonal class $\delta \in H^d(M \times M; \mathbb{Q})$ along the diagonal map $\Delta: M \to M \times M$, evaluate on the fundamental class $[M]$, and compute both sides: the pullback of each Künneth summand $a_i \otimes b_i$ reduces to $a_i \smile b_i$ on $M$, and the evaluation on $[M]$ gives the dual basis pairing $\langle a_i, b_i \rangle = \delta_{ii} = 1$, so the result is the alternating sum of the dimensions of $H^k(M; \mathbb{Q})$. [/proofplan] [step:Pull back the diagonal class along $\Delta$] Let $\Delta: M \to M \times M$ denote the diagonal map $x \mapsto (x, x)$, and let $\pi_1, \pi_2: M \times M \to M$ be the two projections. For any class $\alpha \otimes \beta \in H^*(M; \mathbb{Q}) \otimes H^*(M; \mathbb{Q}) \cong H^*(M \times M; \mathbb{Q})$, we have $\alpha \otimes \beta = \pi_1^*\alpha \smile \pi_2^*\beta$, so \begin{align*} \Delta^*(\alpha \otimes \beta) = \Delta^*(\pi_1^*\alpha \smile \pi_2^*\beta) = (\pi_1 \circ \Delta)^*\alpha \smile (\pi_2 \circ \Delta)^*\beta = \alpha \smile \beta, \end{align*} since $\pi_i \circ \Delta = \operatorname{id}_M$ for $i = 1, 2$. Applying this to the [Diagonal Class](/theorems/2296) $\delta = \sum_i (-1)^{|a_i|} a_i \otimes b_i$: \begin{align*} \Delta^*(\delta) = \sum_i (-1)^{|a_i|} \Delta^*(a_i \otimes b_i) = \sum_i (-1)^{|a_i|} (a_i \smile b_i). \end{align*} [/step] [step:Evaluate on $[M]$ using the dual basis property] The dual basis satisfies $\langle a_i, b_j \rangle = (a_i \smile b_j)[M] = \delta_{ij}$ by definition. Evaluating $\Delta^*(\delta)$ on the fundamental class $[M] \in H_d(M; \mathbb{Q})$: \begin{align*} \Delta^*(\delta)[M] = \sum_i (-1)^{|a_i|} (a_i \smile b_i)[M] = \sum_i (-1)^{|a_i|} \cdot 1 = \sum_i (-1)^{|a_i|}. \end{align*} [/step] [step:Identify the sum with the Euler characteristic] The basis $\{a_i\}$ is a homogeneous basis for $H^*(M; \mathbb{Q}) = \bigoplus_{k=0}^d H^k(M; \mathbb{Q})$. For each degree $k$, the number of basis elements $a_i$ with $|a_i| = k$ equals $\dim_\mathbb{Q} H^k(M; \mathbb{Q})$. Therefore \begin{align*} \sum_i (-1)^{|a_i|} = \sum_{k=0}^d (-1)^k \dim_\mathbb{Q} H^k(M; \mathbb{Q}) = \chi(M), \end{align*} where the last equality is the [Euler Characteristic Equals Homological Euler Characteristic](/theorems/2266) computed over $\mathbb{Q}$. Combining with the previous step: \begin{align*} \Delta^*(\delta)[M] = \chi(M). \end{align*} [/step]