Parametric functions satisfy free theorems which state that they take related arguments to related results. This is formalized by the notion of parametric transformation introduced in section 5 of Logical relations and parametricty: A Reynolds programme for Category Theory and Programming Languages.

In a linear type system, I believe the evaluation function $ev_{AB} : (A \multimap B) \otimes A \to B$ should be parametric because it is defined independently of $A$ and $B$. Indeed, the paper states:

The “$evaluation$” map $ev_{AB} : [A \to B] \times A \to B$ given by $ev(f, x) = f (x)$ is similarly parametric in both $A$ and $B$ (as opposed to just $B$).

The reader will be able to construct similar examples for the internal homs in other closed categories.

However, I wasn't able to see how this result applies to general closed categories. In particular, it seems to fail for vector spaces. Fix some vector space $V$ and consider the evaluation map $ev_A : (A \multimap V) \otimes A \to V$.

Let $k$ be the base field and consider how $ev_k$ and $ev_0$ must be related. In particular, consider the linear relation $R : k \leftrightarrow 0$ which relates every $x \in k$ with $0$. Then I believe any $\varphi \otimes l \in (k \multimap V) \otimes k$ is related by $(R \multimap V) \otimes R$ to $0 \in (0 \multimap V) \otimes 0$. Then by parametricity, $ev$ sends related things to equal things in $V$. It must send $\varphi \otimes l$ to $0$, so it is not evaluation.

It seems like evaluation should still be parametric even when using linear types. Is there good explanation for why the expected free theorem fails to hold?

  • 1
    $\begingroup$ $(R ⊸ V)[φ, ψ] = ∀ s\ t. R(s,t) → φ(s) = ψ(t)$, no? So, $(R ⊸ V)[φ,0] = ∀ s\ t. φ(s) = 0$? I.E. only the $0$ function $k → V$ is related to $0$, and evaluation of the $0$ function gives $0$? $\endgroup$
    – Dan Doel
    Jul 8, 2023 at 16:53

1 Answer 1


Here's an Agda formalization of the non-linear version of your argument, and my comment above:

open import Data.Product renaming (proj₁ to fst; proj₂ to snd)
open import Data.Unit
open import Function
open import Relation.Binary.PropositionalEquality hiding ([_])

  A B C D : Set

ev : (A -> B) × A -> B
ev (f , x) = f x

Rel : Set -> Set -> Set₁
Rel A B = A -> B -> Set

[_⇒_] : Rel A B -> Rel C D -> Rel (A -> C) (B -> D)
[ R ⇒ S ] f g = ∀ a b → R a b -> S (f a) (g b)

[_×_] : Rel A B -> Rel C D -> Rel (A × C) (B × D)
[ R × S ] (w , x) (y , z) = R w y × S x z

module Parametricity
  (pev : ∀{A B C D}
       → (R : Rel A B) (S : Rel C D)
       → [ [ [ R ⇒ S ] × R ] ⇒ S ] ev ev)

  R : Rel A ⊤
  R _ _ = ⊤

  lemma₀ : (φ : A -> B) (g : ⊤ -> B) (x : A)
         → [ R ⇒ _≡_ ] φ g
         → ev (φ , x) ≡ g tt
  lemma₀ φ g x p = pev R _≡_ (φ , x) (g , tt) (p , tt) 

  lemma₁ : ∀(φ : A -> B) (g : ⊤ -> B) → [ R ⇒ _≡_ ] φ g → ∀ x → φ x ≡ g tt
  lemma₁ φ y p x = p x tt tt

The only real difference is that there isn't a unique function $⊤ → A$. But, nevertheless, if we assume that all functions $A → B$ are related to a function $g : ⊤ → B$ via $R → B$, then we get that $\mathsf{ev}(f, x) = g\ \mathsf{tt}$ for arbitrary $x$. Similarly, every function related by $R → B$ to $g$ is constant.

Just because $R$ relates all values of $A$ to the single value of $⊤$ does not mean that $R → B$ relates all functions with type $A → B$ to a given function with type $⊤ → B$. Only the constant functions with matching output are related. In the linear case, this is reduced down to the $0$ function.


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