# Order of products and order of multipliers

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 using the word "multiplier". I think you mean "factor", or perhaps, "multiplicand". –  Zhen Lin Jul 8 '12 at 12:51
Why don't you ask the "more difficult question" there as well? –  Gigili Jul 8 '12 at 12:56
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. –  Asaf Karagila Jul 8 '12 at 12:56
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.