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?



Unicity of Yoneda isomorphism
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. $\mathbf{Set}$.




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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 $\theta\_C$.


