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:
\(\ds \map {\eta_B} f\) | \(=\) | \(\ds \map {\eta_B} {f \circ I_A}\) | Definition of Identity Morphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\eta_B} {\map {\paren {\map {h^A} f} } {I_A} }\) | Definition of Covariant Hom Functor | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\map F f} {\map {\eta_A} {I_A} }\) | 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:
\(\ds \map {\eta_D} {\map {g_*} f}\) | \(=\) | \(\ds \map {\eta_D} {g \circ f}\) | Definition of Postcomposition Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\map F {g \circ f} } u\) |
and:
\(\ds \map F g (\map {\eta_B} f)\) | \(=\) | \(\ds \map {\map F g} {\map {\map F f} u}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\map F {g \circ f} } u\) | 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
\(\ds \map \alpha \eta\) | \(=\) | \(\ds \map {\eta_A} {I_A}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\map F {I_A} } u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {I_{\map F A} } u\) | Definition of Functor | |||||||||||
\(\ds \) | \(=\) | \(\ds u\) | Definition of Identity Mapping |
$\blacksquare$