Given two analytic spaces $X$ and $Y$ over a third analytic space $S$, is it true that
$$
(X\times\_S Y)\_{red}=X\_{red}\times\_{S\_{red}}Y\_{red}?
$$
Any reference is welcome.



Reduction and fibre product
In general the answer is *no*, as shown by the following example.


Given a fibration $f \colon X \to S$, where $X$ and $S$ are smooth $k$-schemes, and setting $Y=\{s\} \hookrightarrow S$ with its reduced scheme structure, then $X\_s:=X \times \_S Y$ is the scheme-theoretic fibre of $f$ over $s$.


Of course such a fibre can be non-reduced even if $X$ and $S$ are both reduced, so in this case $$
(X\times\_S Y)\_{red} \neq X\_{red}\times\_{S\_{red}}Y\_{red}=X \times \_S Y,
$$
because the first space is reduced, whereas the second is not. 


For instance, take $$X= \mathrm{Spec} \, k[x, \, y, \, t]/(ty-x^2), \quad S=\mathrm{Spec}\, k[t]$$
and let $f \colon X \to \mathbb{A}^1$ be the fibration in conics induced by the inclusion $$k[t] \to k[x,\, y,\, t]/(ty-x^2).$$ If $Y= \{0\} \hookrightarrow \mathbb{A}^1$ then $$X\times\_\mathbb{A^1} Y= X\_{red} \times\_{\mathbb{A^1}\_{red}} Y\_{red}=k[x,\, y]/(x^2),$$ which is a double line.


