# Bijection in Yoneda Lemma for Covariant Functors

## Contents

## 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 $\operatorname{id}_A$ be its identity morphism.

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

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

- $\alpha : \operatorname{Nat}(h^A, F) \to F(A) : \eta \mapsto \eta_A(\operatorname{id}_A)$
- $\beta : F(A) \to \operatorname{Nat}(h^A, F) : u \mapsto (X \mapsto (f \mapsto (F(f))(u)))$

## Outline of proof

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

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 $\eta_A(\operatorname{id}_A) = \alpha(\eta)$.

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

## Proof

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

### Injectivity

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

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

Then

\(\displaystyle \eta_B(f)\) | \(=\) | \(\displaystyle \eta_B(f \circ \operatorname{id}_A )\) | Definition of Identity Morphism | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \eta_B((h^A(f))(\operatorname{id}_A))\) | Definition of Covariant Hom Functor | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F(f) (\eta_A(\operatorname{id}_A))\) | Definition of Natural Transformation |

which shows that $\eta$ is determined by $\eta_A(\operatorname{id}_A) = \alpha(\eta)$.

$\Box$

### Surjectivity

Let $u \in F(A)$.

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

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

We have to show that the following diagram commutes:

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

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

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

We have:

\(\displaystyle \eta_D(g_*(f))\) | \(=\) | \(\displaystyle \eta_D(g \circ f)\) | Definition of Postcomposition Mapping | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F(g\circ f)(u)\) |

and

\(\displaystyle F(g)(\eta_B(f))\) | \(=\) | \(\displaystyle F(g)(F(f)(u))\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F(g \circ f)(u)\) | Definition of Covariant Functor, applied to $F$ |

$\Box$

### Reverse bijections

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

It remains to show that $\alpha(\beta(u)) = u$ for all $u \in F(A)$.

Let $\eta = \beta(u)$.

Then

\(\displaystyle \alpha(\eta)\) | \(=\) | \(\displaystyle \eta_A(\operatorname{id}_A)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle F(\operatorname{id}_A)(u)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \operatorname{id}_{F(A)}(u)\) | Definition of Functor | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle u\) | Definition of Identity Mapping |

$\blacksquare$