Proof (a).
\\(K_B\\) is the claimed constant functor which maps every set to \\(B\\) and every morphism to \\(Id_B\\).
To prove that \\(K_B\\) is indeed a functor, it suffices to show that for all sets \\(S\\), the mapping \\(K_S\\) satisfies the two functor laws.
The identity law for \\(K_S: Set \rightarrow Set\\) says that for every \\(A \in Ob(Set)\\), we have that \\(K_S(Id_A) = Id_{K_S(A)}\\). Since \\(K_S(A) = S\\), this requirement restates to \\(K_S(Id_A) = Id_S\\).
This is certainly true, as \\(K_S\\) maps any morphism whatsoever to \\(Id_S\\).
The composition law requires that for composable morphisms \\(f, g\\), we have that \\(K_S(f \triangleright g) = K_S(f) \triangleright K_S(g)\\). That's equivalent to saying that \\(Id_S = Id_S \triangleright Id_S\\), which is an axiomatic fact.
\\(K_B\\) is the claimed constant functor which maps every set to \\(B\\) and every morphism to \\(Id_B\\).
To prove that \\(K_B\\) is indeed a functor, it suffices to show that for all sets \\(S\\), the mapping \\(K_S\\) satisfies the two functor laws.
The identity law for \\(K_S: Set \rightarrow Set\\) says that for every \\(A \in Ob(Set)\\), we have that \\(K_S(Id_A) = Id_{K_S(A)}\\). Since \\(K_S(A) = S\\), this requirement restates to \\(K_S(Id_A) = Id_S\\).
This is certainly true, as \\(K_S\\) maps any morphism whatsoever to \\(Id_S\\).
The composition law requires that for composable morphisms \\(f, g\\), we have that \\(K_S(f \triangleright g) = K_S(f) \triangleright K_S(g)\\). That's equivalent to saying that \\(Id_S = Id_S \triangleright Id_S\\), which is an axiomatic fact.
Version 2.1.8p2
