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Constant functors.
Let \(C\) and \(D\) be categories. Given any object \(d\) in \(D\), we can define the constant functor \(K_d: C \rightarrow D\) on \(d\). This functor sends every object \(C\) to \(d \in Ob\ D\), and every morphism of \(C\) to the identity morphism on \(d\).
(a) Take the set \(B\) = {\(T,F\)}. Show that the constant functor \(K_B: Set \rightarrow Set\) obeys the two functor laws: preservation of composition and preservation of identities.
(b) Implement in Haskell the constant functor on the type Bool.
Comments
Edit: I was confused because we went from a constant functor from Set_Bool -> Set_Bool to a constant functor from Type -> Type. I get now that the constant that they are looking for is the \(Bool \in Obj_{Hask} \)
> In b: am confused by what it means by "the constant functor" aren't there two constant functors? Edit: I was confused because we went from a constant functor from Set_Bool -> Set_Bool to a constant functor from Type -> Type. I get now that the constant that they are looking for is the \\(Bool \in Obj_{Hask} \\)I just noticed this was written in "draft" and I had forgotten to post it :P
I just noticed this was written in "draft" and I had forgotten to post it :P > newtype ConstF c a = Const c > instance Functor (ConstF c) where > fmap f (Const c) = Const c > kb :: ConstF Bool a > kb = ConstF True > -- fmap id (Const c) = id (Const c) > -- Const c = Const c <-- applying the definition of fmap > -- fmap (f.g) (Const c) = (fmap f . fmap g) (Const c) > -- Const c = fmap f (fmap g (Const c)) <<- applying the definition of fmap on the left side > -- Const c = fmap f (Const c) <-- applying the definition of fmap on the right side > -- Const c = Const cProof (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.
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.