# Bijection in Yoneda Lemma for Covariant Functors

## Theorem

Let $C$ be a locally small category.

Let $\mathbf {Set}$ be the category of sets.

Let $F: C \to \mathbf {Set}$ be a covariant functor.

Let $A \in C$ be an object.

Let $I_A$ be its identity morphism.

Let $h^A = \map {\operatorname {Hom} } {A, -}$ be its covariant hom-functor.

The class of natural transformations $\map {\operatorname {Nat} } {h^A, F}$ is a small class, and:

$\alpha: \map {\operatorname {Nat} } {h^A, F} \to \map F A: \eta \mapsto \map {\eta_A} {I_A}$
$\beta: \map F A \to \map {\operatorname {Nat} } {h^A, F}: u \mapsto \paren {X \mapsto \paren {f \mapsto \map {\paren {\map F f} } u} }$

## Outline of proof

The crucial point is that the role of morphisms $f \in \map {h^A} B = \map {\operatorname {Hom} } {A, B}$ is close to their role under the hom functor $h^A$: We have $f = f \circ I_A = \map {\paren {\map {h^A} f} } {I_A}$.

Natural transformations $\eta: h^A \to F$ translate what happens with $f$ under $h^A$ to something involving only $F$, so that $\eta$ is determined by $\map {\eta_A} {I_A} = \map \alpha \eta$.

Making the calculation explicit, we know what the reverse bijection $\beta$ should be.

## Proof

The fact that $\map {\operatorname {Nat} } {h^A, F}$ is a small class follows when we prove that the mappings are bijections.

### Injectivity

Let $\eta \in \map {\operatorname {Nat} } {h^A, F}$ and $B \in C$ an object.

Let $f \in \map {h^A} B = \map {\operatorname {Hom} } {A, B}$ be a morphism.

Then:

 $\displaystyle \map {\eta_B} f$ $=$ $\displaystyle \map {\eta_B} {f \circ I_A}$ Definition of Identity Morphism $\displaystyle$ $=$ $\displaystyle \map {\eta_B} {\map {\paren {\map {h^A} f} } {I_A} }$ Definition of Covariant Hom Functor $\displaystyle$ $=$ $\displaystyle \map {\map F f} {\map {\eta_A} {I_A} }$ Definition of Natural Transformation

which shows that $\eta$ is determined by $\map {\eta_A} {I_A} = \map \alpha \eta$.

$\Box$

### Surjectivity

Let $u \in \map F A$.

Inspired by the result above, we define $\map \beta u = \eta$ by $\map {\eta_B} f = \map {\map F f} u$ for $B \in C$ and $f \in \map {\operatorname {Hom} } {A, B}$.

Let $B, D \in C$ and $g: B \to D$ be a morphism.

We have to show that the following diagram commutes:

$\xymatrix{ \map {\operatorname {Hom} } {A, B} \ar[d]^{\eta_B} \ar[r]^{g_*} & \map {\operatorname {Hom} } {A, D} \ar[d]^{\eta_D} \\ F(B) \ar[r]^{\map F g} & \map F D}$

where $g_* = \map {\operatorname {Hom} } {A, g}$ is the postcomposition mapping.

Let $f \in \map {\operatorname {Hom} } {A, B}$.

We have:

 $\displaystyle \map {\eta_D} {\map {g_*} f}$ $=$ $\displaystyle \map {\eta_D} {g \circ f}$ Definition of Postcomposition Mapping $\displaystyle$ $=$ $\displaystyle \map {\map F {g \circ f} } u$

and:

 $\displaystyle \map F g (\map {\eta_B} f)$ $=$ $\displaystyle \map {\map F g} {\map {\map F f} u}$ $\displaystyle$ $=$ $\displaystyle \map {\map F {g \circ f} } u$ Definition of Covariant Functor, applied to $F$

$\Box$

### Reverse bijections

By construction, $\map \beta {\map \alpha \eta} = \eta$ for all $\eta \in \map {\operatorname {Nat} } {h^A, F}$.

It remains to show that $\map \alpha {\map \beta u} = u$ for all $u \in \map F A$.

Let $\eta = \map \beta u$.

Then

 $\displaystyle \map \alpha \eta$ $=$ $\displaystyle \map {\eta_A} {I_A}$ $\displaystyle$ $=$ $\displaystyle \map {\map F {I_A} } u$ $\displaystyle$ $=$ $\displaystyle \map {I_{\map F A} } u$ Definition of Functor $\displaystyle$ $=$ $\displaystyle u$ Definition of Identity Mapping

$\blacksquare$