# Are there maps that take ordered pairs to ordered pairs? [closed]

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

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

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• Yes... of course – JMoravitz Jun 6 at 13:29
• Yes, why not?.. – ArsenBerk Jun 6 at 13:29
• 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. – JMoravitz Jun 6 at 13:31
• There is no such map if $S'=\emptyset$ and $S\ne \emptyset$. In all other cases, at least one such map exists. – Hagen von Eitzen Jun 6 at 13:34

An easy, concrete example. Consider the identity function on $$S \times S$$.