Euler characteristic of a quotient space

I have a question relating to an answer on MathOverflow.net. The cited answer says:

Let $X$ be a topological space for which [the Euler characteristic] $\chi(X)$ is defined and behaves in the expected way for unions, Cartesian products, and quotients by a finite free action. ... [Then] $$\chi(X^{(2)}) = \frac{\chi(X \times X) - \chi(\operatorname{Diag}(X))}{2} + \chi(X) = \frac{\chi(X)^2 + \chi(X)}{2}$$ [where $X^{(2)}$ denotes the symmetric square of $X$].

Question: Does anyone know a reference for this result, or, failing that, a short proof? For the application that I have in mind I need the result for algebraic varieties over an algebraically closed field (whose characteristic may be positive), but a more general result would be nice to see.

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The Cartesian product is the union of the diagonal and its complement, and a finite group $C_2$ acts freely on that complement with quotient the symmetric square minus a copy of $X$, so $$\chi(X \times X) = \chi(\text{Diag}(X)) + 2 (\chi(X^{(2)}) - \chi(X))$$
and the conclusion follows. (This is essentially combinatorics; run through the argument for $X$ a finite discrete space if this part is unclear.)
Thanks! I believe a slight generalisation of the statement should also be true, namely $\chi(\mathscr{O}_{X\times X}) = \chi(\mathscr{O}_{\operatorname{Diag}(X)}) + 2(\chi(\mathscr{O}_{X^{(2)}}) - \chi(\mathscr{O}_X))$ where $X$ is an algebraic variety and $\chi$ is the Euler characteristic of the sheaf cohomology. Does this generalisation follow easily (I don't immediately see how), or is it much more complicated? –  Hamish Aug 13 '12 at 17:07
That was my first thought too. If $X = \operatorname{Spec}(A)$ is affine for example, then the only candidate sequence I can think of is $0 \rightarrow (A \otimes A)^{C_2} \rightarrow A \otimes A \stackrel{\phi}{\rightarrow} B$ where $B = \bigoplus_{g\in C_2} (A \otimes A) = (A \otimes A) \oplus (A\otimes A)$ and $\phi(a\otimes b) = (0, b\otimes a - a \otimes b)$. But this is not surjective. I think I'll ask it as a new question. –  Hamish Aug 13 '12 at 19:20