1

I have a problem understanding how Yoneda Lemma gives a natural isomorphism between functors. Let $F,G \colon \mathcal{A}\to \mathcal{B}$ two functors and I have a bifunctorial isomorphism $\mathrm{Hom}(F(\cdot), \cdot) \cong \mathrm{Hom}(G(\cdot), \cdot)$. I want to prove that $F\cong G$.

For $A \in \mathcal{A}$ I have a functorial isomorphism $\mathrm{Hom}(F(A), \cdot) \cong \mathrm{Hom}(G(A), \cdot)$, and the Yoneda Lemma implies that $$\operatorname{Nat}(\mathrm{Hom}(F(A), \cdot), \, \mathrm{Hom}(G(A),\cdot)) \cong \mathrm{Hom}(G(A), F(A)) \,.$$ I will denote $\eta \colon \mathrm{Hom}(F(A), \cdot) \xrightarrow{\sim} \mathrm{Hom}(G(A), \cdot)$ the previous isomorphism, and then a natural transformation $\mathrm{Hom}(F(A),\cdot) \cong \mathrm{Hom}(G(A), \cdot)$ gives a unique mapping $G(A) \to F(A)$ given by $\eta_{F(A)}(\mathrm{id}_{F(A)})$. How I can prove that this collection of morphisms are a natural isomorphism $F \cong G$? (i.e., $\eta_{F(A)}(\mathrm{id}_{F(A)})$ is an isomorphism for every $A$ and these morphisms are naturals).

Jendrik Stelzner
  • 17,081
user34977
  • 191
  • 5

1 Answers1

3

We have natural isomorphisms (bijections) $$\alpha_{X,Y} : \hom(F(X),Y) \xrightarrow{~\sim~} \hom(G(X),Y)$$ for $X \in \mathcal{A}$, $Y \in \mathcal{B}$. If we fix $X$, these yield a natural isomorphism of functors $$\alpha_{X,-} : \hom(F(X),-) \xrightarrow{~\sim~}\hom(G(X),-).$$ Recall that the Yoneda Lemma implies that every (iso)morphism of functors $\hom(Y,-) \to \hom(Y',-)$ is induced by a unique (iso)morphism $Y' \to Y$. Thus, the isomorphism above is induced by an isomorphism $$\beta_X : G(X) \xrightarrow{~\sim~} F(X).$$ Here, $X$ was arbitrary. Now it is straight forward to check that, since $\alpha$ is natural in $X$, these isomorphisms $\beta$ are natural in $X$ as well.

Here are more details (but really, this is just a matter of chasing down the definitions): By construction of $\beta$ and the Yoneda construction, we have $$\alpha_{X,Y}(f) = f \circ \beta_X$$ for all $f : F(X) \to Y$. If $h : X \to X'$ is a morphism, naturality of $\alpha$ with respect to $h$ says $$G(h)^* \circ \alpha_{X',Y} = \alpha_{X,Y} \circ F(h)^*.$$ Evaluating this at the identity $\mathrm{id}_{F(X')}$, we get on the left hand side $$G(h)^*(\alpha_{X',Y}(\mathrm{id}_{F(X')})) = G(h)^*(\beta_{X'}) = \beta_{X'} \circ G(h),$$ and on the right hand side $$\alpha_{X,Y}(F(h)^*(\mathrm{id}_{F(X')})) = \alpha_{X,Y}(F(h)) = F(h) \circ \beta_X.$$ Thus, $\beta$ is natural.

More generally, any morphism of functors $\hom(F(-),-) \to \hom(G(-),-)$ is induced by a unique morphism of functors $G \to F$.

Martin Brandenburg
  • 181,922