Let $\mathcal{A}\to\textrm{Sets}$ be a covariant functor, $X, X'\in\operatorname{Obj}(\mathcal{A})$ objects, and $(X, \Phi), (X', \Phi')$ be representations of $F$, i.e. $\Phi: F\to\operatorname{Hom}_{\mathcal{A}}(X, \cdot)$ and $\Phi': F\to\operatorname{Hom}_{\mathcal{A}}(X', \cdot)$ are natural equivalences. Show that there is a unique isomorphism $f: X\to X'$ such that for any object $Y\in\operatorname{Obj}(\mathcal{A})$: $$\Phi_Y = \operatorname{Hom}_{\mathcal{A}}(\cdot, Y)(f)\circ \Phi_Y'$$

What I did so far:

$\Phi'\circ \Phi^{-1}: \operatorname{Hom}_{\mathcal{A}}(X, \cdot)\to\operatorname{Hom}_{\mathcal{A}}(X', \cdot)$

$\Phi\circ {\Phi'}^{-1}: \operatorname{Hom}_{\mathcal{A}}(X', \cdot)\to\operatorname{Hom}_{\mathcal{A}}(X, \cdot)$

are both (isomorphic) natural transformations. The Lemma of Yoneda states the bijectivity of the following mappings:

$$v: \operatorname{Nat}(\operatorname{Hom}_{\mathcal{A}}(X, \cdot), \operatorname{Hom}_{\mathcal{A}}(X', \cdot))\to\operatorname{Hom}_{\mathcal{A}}(X, X')$$ $$\phi\mapsto\phi_X(\textrm{Id}_X)$$ and $$v': \operatorname{Nat}(\operatorname{Hom}_{\mathcal{A}}(X', \cdot), \operatorname{Hom}_{\mathcal{A}}(X, \cdot))\to\operatorname{Hom}_{\mathcal{A}}(X', X)$$ $$\phi\mapsto\phi_{X'}(\textrm{Id}_{X'})$$

Now I could guess the morhpism that I search for is $f := v'(\Phi\circ {\Phi'}^{-1})$. In order to show that $f$ is an isomorphism, a good candidate for its inverse could be $\tilde{f} := v(\Phi'\circ \Phi^{-1})$.

So first I need to show that indeed $f\circ\tilde{f}=\textrm{Id}_X$. Afterwards I would like to show $$\forall Y\in\operatorname{Obj}(\mathcal{A}): \Phi_Y = \operatorname{Hom}_{\mathcal{A}}(\cdot, Y)(f)\circ \Phi_Y'$$ Finally I still need to show the uniqueness of $f$. So far I haven't really used the Lemma of Yoneda (only as a hint how to construct $f$).

I don't know whether this is the correct way. Any help is appreciated, especially hints that are easy to understand rather than complete abstract solutions.

  • 1
    $\begingroup$ The fact that $f$ and $\tilde{f}$ (you defined) are inverse of each other is clear: $f$ is given as the preimage by the Yoneda embedding of $\phi^{'} \circ \phi^{-1}$ and $\tilde{f}$ as the preimage of $\phi \circ \phi^{'-1}$. Thus $f\circ \tilde{f}$ is given as the preimage of $Id_{Hom(X,-)}$ if you see $f\circ \tilde{f}$ as an arrow in $\mathcal{A}$ and the compositions of $\phi$ and $\phi^{'}$ as arrows in $Fun(\mathcal{A},Set)$. $\endgroup$ – Dominique Mattei Oct 9 at 16:43
  • $\begingroup$ Thanks Dominique for your comment. This is not clear to me. I should mention that the version of the yoneda lemmathat I had in my lecture only states that the natural transformations from $\operatorname{Hom}_{\mathcal{A})(X, \cdot)$ to some functor $F: \mathcal{A}\to\textrm{Sets}$ are in bijection with the set $F(X)$, where the bijection is like those mappings in my question. Sorry, don't ask me why the formatting doesn't work. $\endgroup$ – S. M. Roch Oct 9 at 19:21
  • $\begingroup$ The classical Yoneda lemma is the one you stated but it also tells that the bijection $Nat(Hom(X,-), F) \to F(X)$ is natural in both $X$ and $F$, but maybe that is what your exercice tries to prove. To me, the most easy way to think of Yoneda lemma is "The functor $X \mapsto Hom(X,-)$ from $\mathcal{A}^{op}$ to $Fun(\mathcal{A},Set)$ is fully faithful (that is, an embedding)". $\endgroup$ – Dominique Mattei Oct 10 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.