The Yoneda Lemma
2.3.2 The inverse Instead of taking a transformation Φ and providing an element α (Φ) ∈ F ( A ), we now take an element x ∈ F ( A ) and provide a natural transformation α − 1 ( x ) : Hom A → F . Since we want α − 1 ( x ) to be a natural transformation, it must also satisfy the naturality condition, so we want the equation
( α − 1 ( x ))
X ( f ) = F ( f )( x )
to hold. This is simply a substitution into the equation derived above, replacing α (Φ) with x and Φ with α − 1 ( x ). As discussed above, this equation is sufficient to specify α − 1 . Now it remains only to show that α − 1 ◦ α = id F ( A ) and α ◦ α − 1 = id Nat(Hom A,F ) .
For the first equation (a property known as being left inverse ) we observe that
( α − 1 ◦ α )( x ) = α ( α − 1 ( x )) = α − 1 ( x )
A (id A )
= F (id A )( x ) = id F ( A )( x ) = x
Thus for any element x we have ( α − 1 ◦ α )( x ) = x , the definition of the identity morphism in Set (from Definition 2). The proof that α − 1 is right inverse to α is slightly trickier, as we have a natural transformation and must show that each component is equal.
( α ◦ α − 1 )(Φ) = α − 1 ( α (Φ)) = α − 1 (Φ
A (id A ))
Taking the X component for any X and applying it to any f yields
( α − 1 (Φ
A (id A ))) X ( f ) = F ( f )(Φ A (id A )) = (Φ A ◦ F ( f ))(id A )
= (Hom A ( f ) ◦ Φ X )(id A ) = Φ X (Hom A ( f )(id A )) = Φ X (id A ◦ f ) = Φ X ( f )
Since this holds for all X and f we then have α − 1 ( α (Φ)) = Φ so α ◦ α − 1 = id Nat(Hom A,F ) , again using the definition of identity function from Definition 2. Thus we have shown that α − 1 is both left and right inverse to α , so we do indeed have an isomorphism as required.
2.4 Naturality The previous section of the proof involved several nested levels of proving something for all values of a variable, whether f , X , or Φ. Now we will also allow F and A to take different values. We seek to show that there is a natural isomorphism between the functor LHS defined by LHS ( A,F ) = Nat(Hom A ,F ) and the application functor defined by RHS ( A,F ) = F ( A ). The astute reader may guess that the components of this natural isomorphism will be given by the isomorphisms α defined above. In fact, since α depended
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