Also here's a [diagram of the commutative square](https://tikzcd.yichuanshen.de/#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) in puzzle 170 (similar to the one drawn by John above).

Because the diagram commutes, we have \$$(f,g) = (f,1_{y'}) \circ (1_x,g) = (1_{x'},g) \circ (f,1_y) \$$. If \$$(f,g) \$$ is a preorder on \$$X^\mathrm{op} \times Y\$$, then the diagram is the solution to puzzle 169. You can translate the category theoretic concepts back into preorder ones for a proof: If \$$x' \le x\$$ in \$$X\$$, then \$$x \le x'\$$ in \$$X^\mathrm{op}\$$ and if \$$y \le y'\$$ in \$$Y\$$, then by reflexitivity, we have \$$(x,y) \le (x',y)\$$ and \$$(x',y) \le (x',y')\$$. By transitivity, we have \$$(x,y) \le (x',y)\$$ and \$$(x',y) \le (x',y')\$$ implies \$$(x,y)\le (x',y')\$$.