# Unicity of Yoneda isomorphism

I am wondering if there is only one unique Yoneda isomorphism, that is a natural isomorphism (natural in C and P, that is) between Hom(yC,P) and PC.

The Yoneda lemma says that there exists at least one, and its common proof is constructive, so we have an example. But is it the only one?

If it is unique, how do we prove that? If not, is there any counter-example?

If there is an isomorphism $\mathrm{Hom}(y C, P) \cong P (C)$ natural in $P$ and $C$, then by restricting to the representable functors one obtains an automorphism of the identity functor of the category $C$ comes from. There are categories for which the identity functor has a non-trivial automorphism group, e.g. $\mathbf{Ab}$, and also categories for which the identity functor has a trivial automorphism group, e.g. $\mathbf{Set}$.
In more detail: let $P = y D$. Then $\mathrm{Hom}(y C, y D) \cong \mathcal{T}(C, D)$, and if we compose with the standard Yoneda isomorphism we get an isomorphism $y D \cong y D$, natural in $D$. The Yoneda embedding $y : \mathcal{T} \to [\mathcal{T}^\mathrm{op}, \mathbf{Set}]$ is fully faithful, so this determines an automorphism of the identity functor on $\mathcal{T}$.
Conversely, suppose $\theta : \mathrm{id}_{\mathcal{T}} \Rightarrow \mathrm{id}_{\mathcal{T}}$ is an automorphism. Then we can compose it with the Yoneda embedding and use functoriality of $\mathrm{Hom}$ to get a new natural bijection $\mathrm{Hom}(y C, P) \cong P (C)$: explicitly, it is the map defined by evaluating $P$ at $\theta_C$.
• Hi Zhen Lin, thank you for your answer. However, I don't see clearly the automorphism: Let be $P$ a representable functor with $P=yD=\mathrm{Hom}(C,D)$. We then have that any Yoneda isomorphism is a natural isomorphism $\mathrm{Hom}(yC,yD)\cong\mathrm{Hom}(C,D)$. Then an automorphism of identity applied to $P$ is something like an isomorphic natural transformation $P\rightarrow \theta(P)$, thus $\theta\cong 1_\mathbf{T}$ (where $\mathbf{T}$ is the category to which $P$ belongs). But then I don't see why we supposed $P$ representable? Aug 24 '13 at 14:46
• @AlmeoMaus When you say "an automorphism of identity applied to P is something like an isomorphic natural transformation $P\rightarrow \theta(P)$" it sounds like you are forgetting that a natural transformation is natural between two functors, not "natural" at each value. Aug 24 '13 at 18:56
• The high concept answer is that $[\mathcal{T}^\mathrm{op}, \mathbf{Set}]$ is the free cocompletion of $\mathcal{T}$ when $\mathcal{T}$ is small, so the automorphism group of its identity functor (which is what you are secretly asking about) is isomorphic to the automorphism group of the identity functor of $\mathcal{T}$. Aug 25 '13 at 0:03