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Let $f$ and $g$ are binary relations (on some set $\mho$). The function $f\times^{C} g$ is defined by the formula: $(f\times^{C} g) a = g\circ a \circ f^{-1}$ (for every binary relation $a$ on $\mho$.

Suppose $f$ and $g$ are non-empty. Can we restore $f$ and $g$ knowing only the value of $f\times^{C} g$?

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  • $\begingroup$ When you say, "knowing only the value of $f \times^C g$", do you mean you can feed any binary relation into this construction and you'll know the resulting binary relation? $\endgroup$
    – Hugh Denoncourt
    Commented Jul 5, 2012 at 21:37
  • $\begingroup$ @Hugh Denoncourt: We should feed only values arising as a $\times^C$ product of some binary relations $f$ and $g$ (It seems that there are values which $f\times^C g$ cannot take. We don't consider these values.) Once more: We know $f\times^C g$ and that it is non-empty. We need to infer $f$ and $g$. $\endgroup$
    – porton
    Commented Jul 5, 2012 at 21:47
  • $\begingroup$ A more difficult question continuing this trend: mathoverflow.net/questions/101576/… $\endgroup$
    – porton
    Commented Jul 7, 2012 at 18:56

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Let $a_{u,v} = \{(u,v)\}$. Then $g\circ a_{u,v}\circ f^{-1} = f^{-1}(u)\times g(v)$. Assuming $g$ is nonempty, pick $v\in \text{dom}(g)$. Then $\text{dom} \left[(f\times^C g)a_{u,v}\right] = f^{-1}(u)$, so varying $u$ allows you to recover $f$. Similarly picking $u\in\text{range}(f)$ and varying $v$ you can recover $g$.

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