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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.

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    $\begingroup$ The left is reduced, and the right need not be -- say $X,Y$ are two tangent curves in the plane $S$. $\endgroup$ Apr 20 '17 at 10:44
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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.

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