1. The products of identity morphism for pairs of objects, one from \$$\mathcal {C}\$$ and the other from \$$\mathcal {D}\$$.

2. Because composition in each multiplicand is associative.

3. It has objects \$$(1,a)\$$, \$$(1,b)\$$, and a single non-identity morphism: \$$(\mathrm{id}_1,f):(1,a)\to(1,b)\$$.

4. It is the product preorder on the Cartesian product of the underlying sets (since \$$(f,g):(c,d)\to(c',d')\$$ exists iff there is a morphism from \$$c\$$ to \$$c'\$$ and a morphism from \$$d\$$ to \$$d'\$$, i.e., products are ordered iff both their multiplicands are (as in Example 1.47)).