Anindya - right! There are always two ways to "factorize" an inequality in a product of preorders, and as Keith noted, this is a special case of a more general fact: there are always two ways to factorize a morphism in a product of categories.

Namely, as you said a while back, any morphism in a product of categories \\(\mathcal{C} \times \mathcal{D}\\) is of the form

\[ (f,g) : (x,y) \to (x',y') \]

and we always have

\[ (f,g) = (f,1_{y'}) \circ (1_x, g) \]

but also

\[ (f,g) = (1_{x'}, g) \circ (f, 1_y) \]

It's nice to think of this using a commutative square:

\[

\begin{matrix}

& & (f, 1_y) & & \\\\

& (x,y) & \rightarrow & (x',y) &\\\\

(1_x, g) & \downarrow & & \downarrow & (1_{x'},g) \\\\

& (x,y') & \rightarrow & (x',y') &\\\\

& & (f,1_{y'}) & & \\\\

\end{matrix}

\]

The morphism \\((f,g)\\) goes down the diagonal of this square, from upper left to lower right - but this typesetting system is too wimpy for me to draw it here!

Namely, as you said a while back, any morphism in a product of categories \\(\mathcal{C} \times \mathcal{D}\\) is of the form

\[ (f,g) : (x,y) \to (x',y') \]

and we always have

\[ (f,g) = (f,1_{y'}) \circ (1_x, g) \]

but also

\[ (f,g) = (1_{x'}, g) \circ (f, 1_y) \]

It's nice to think of this using a commutative square:

\[

\begin{matrix}

& & (f, 1_y) & & \\\\

& (x,y) & \rightarrow & (x',y) &\\\\

(1_x, g) & \downarrow & & \downarrow & (1_{x'},g) \\\\

& (x,y') & \rightarrow & (x',y') &\\\\

& & (f,1_{y'}) & & \\\\

\end{matrix}

\]

The morphism \\((f,g)\\) goes down the diagonal of this square, from upper left to lower right - but this typesetting system is too wimpy for me to draw it here!