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)).

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)).