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Specifically, suppose we have two sets, $S$ and $S'$. Is there a mapping $f: S \times S \rightarrow S' \times S'$?

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closed as off-topic by Ennar, Lee David Chung Lin, Leucippus, Cesareo, Shogun Jun 7 at 18:12

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    $\begingroup$ Yes... of course $\endgroup$ – JMoravitz Jun 6 at 13:29
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    $\begingroup$ Yes, why not?.. $\endgroup$ – ArsenBerk Jun 6 at 13:29
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    $\begingroup$ You can have a mapping from any nonempty set $A$ to any nonempty set $B$, regardless how complicated of a mess $A$ or $B$ happen to be. Remember that a set of ordered pairs is still a set, even if it is "more complicated" than more vanilla sets that you are more comfortable with. $\endgroup$ – JMoravitz Jun 6 at 13:31
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    $\begingroup$ There is no such map if $S'=\emptyset$ and $S\ne \emptyset$. In all other cases, at least one such map exists. $\endgroup$ – Hagen von Eitzen Jun 6 at 13:34
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An easy, concrete example. Consider the identity function on $S \times S$.

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