[proofplan]
We compute $\chi(X \times Y)$ using the CW structure on the product, where a $p$-cell of $X$ and a $q$-cell of $Y$ combine to give a $(p+q)$-cell of $X \times Y$. The alternating sum of cell counts factors as a Cauchy product of two finite sums, yielding $\chi(X) \cdot \chi(Y)$.
[/proofplan]
[step:Describe the product CW structure and count cells by dimension]
The product $X \times Y$ inherits a CW structure in which the cells are products of cells of $X$ and $Y$: for each $p$-cell $e_\alpha$ of $X$ and each $q$-cell $f_\beta$ of $Y$, the product $e_\alpha \times f_\beta$ is a $(p + q)$-cell of $X \times Y$.
Let $c_k(Z)$ denote the number of $k$-cells of a CW complex $Z$. The number of $n$-cells of $X \times Y$ is
\begin{align*}
c_n(X \times Y) = \sum_{p + q = n} c_p(X) \cdot c_q(Y).
\end{align*}
[guided]
The standard CW structure on the product of two CW complexes assigns to each pair consisting of a $p$-cell $e_\alpha$ of $X$ and a $q$-cell $f_\beta$ of $Y$ a single $(p+q)$-cell $e_\alpha \times f_\beta$ of $X \times Y$. (This requires working in the category of compactly generated spaces to ensure the product topology agrees with the CW topology, but for finite CW complexes there is no issue.)
The count of $n$-cells in $X \times Y$ is obtained by summing over all ways to decompose $n$ as $p + q$ with $p, q \geq 0$, and for each such decomposition, the number of cell pairs is $c_p(X) \cdot c_q(Y)$.
[/guided]
[/step]
[step:Factor the alternating sum as a Cauchy product to obtain $\chi(X) \cdot \chi(Y)$]
The Euler characteristic of $X \times Y$ is
\begin{align*}
\chi(X \times Y) &= \sum_{n \geq 0} (-1)^n c_n(X \times Y) = \sum_{n \geq 0} (-1)^n \sum_{p + q = n} c_p(X) \cdot c_q(Y).
\end{align*}
Since $(-1)^n = (-1)^{p+q} = (-1)^p (-1)^q$, we may write
\begin{align*}
\chi(X \times Y) = \sum_{p, q \geq 0} (-1)^p c_p(X) \cdot (-1)^q c_q(Y).
\end{align*}
Both sums are finite (since $X$ and $Y$ are finite CW complexes), so the double sum factors:
\begin{align*}
\chi(X \times Y) = \left(\sum_{p \geq 0} (-1)^p c_p(X)\right) \cdot \left(\sum_{q \geq 0} (-1)^q c_q(Y)\right) = \chi(X) \cdot \chi(Y).
\end{align*}
[guided]
We substitute the cell count formula from the previous step:
\begin{align*}
\chi(X \times Y) = \sum_{n \geq 0} (-1)^n \sum_{p + q = n} c_p(X) \cdot c_q(Y).
\end{align*}
The sign $(-1)^n = (-1)^{p+q}$ splits as a product $(-1)^p \cdot (-1)^q$, so the double sum becomes
\begin{align*}
\chi(X \times Y) = \sum_{p, q \geq 0} (-1)^p c_p(X) \cdot (-1)^q c_q(Y).
\end{align*}
This is a product of two finite sums (finiteness is guaranteed because $X$ has finitely many cells, so $c_p(X) = 0$ for $p$ large enough, and likewise for $Y$). By the distributive law for finite sums:
\begin{align*}
\sum_{p, q \geq 0} a_p b_q = \left(\sum_p a_p\right)\left(\sum_q b_q\right),
\end{align*}
and setting $a_p = (-1)^p c_p(X)$ and $b_q = (-1)^q c_q(Y)$ gives $\chi(X \times Y) = \chi(X) \cdot \chi(Y)$.
Note that this argument uses only the cell-count definition of $\chi$, which equals the homological Euler characteristic by the [Euler Characteristic Equals Homological Euler Characteristic](/theorems/2266) theorem.
[/guided]
[/step]
