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I asked this question (and have received an answer) at MathOverflow.

Now a little more difficult question:

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 $f_0$, $f_1$, $g_0$, and $g_!$ are non-empty. Knowing $f_0\times^{C} g_0\subseteq f_1\times^{C} g_1$, can we infer $f_0\subseteq f_1$ and $g_0\subseteq g_1$?

I am especially interested in short elegant proofs, because I am working on generalizing this.

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    You keep posting questions, requesting *us* to give *you* short and elegant proofs for things you wish to generalize. Either generalize or prove it for yourself. Moreover in many of the cases you post your own answer within the hour after posting here. I see a pattern here, and it seems that you're not trying hard enough before posting questions.2012-07-08

1 Answers 1

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Let $a=\{x\}\times\mho$ for some $x\in\mho$. Then $(f\times^C g)a = g\circ(\{x\}\times\mho)\circ f^{-1} = f[\{x\}]\times g[\mho]$.

Thus $f_0[\{x\}]\times g_0[\mho] \subseteq f_1[\{x\}]\times g_1[\mho]$.

$g_0[\mho]\ne\varnothing$ and $g_1[\mho]\ne\varnothing$ because $g_0$ and $g_1$ are non-empty. Thus by properties of Cartesian product $f_0[\{x\}]\subseteq f_1[\{x\}]$ (for every $x\in\mho$). From this follows $f_0\subseteq f_1$.

$g_0\subseteq g_1$ is similar.