4
$\begingroup$

Emily Riehl's "Category Theory in Context, ${\rm Exercise}~2.2.{\rm ii}.$

Explain why the Yoneda lemma does not dualize to classify natural transformations from an arbitrary set-valued functor to a represented functor.

After reading a related question I finally understand what the question is; however, I still do not know how to answer it. For reference, the Yoneda lemma is stated as

For any functor $F:{\rm C}\to{\rm Set}$, whose domain ${\rm C}$ is locally small and any object $c\in{\rm C}$, there is a bijection $${\rm Hom}({\rm C}(c,-),F)\cong Fc$$ that associates a natural transformation $\alpha:{\rm C}(c,-)\Rightarrow F$ to the element $\alpha_c(1_c)\in Fc$. Moreover, this correspondence is natural in both, $c$ and $F$.

The only thoughts I have had were about size issues, i.e. that ${\rm Hom}(F,C(c,-))$ fails to be a set assuming an arbitrary functor, or that it might be problematic to define an actual bijection $\beta:F\to{\rm C}(c,-)$. Both do not seem right and I have got the feeling that I am missing something essential.

Could you please explain to me what the crux of the exercise is, i.e. what the expected answer is?

Thanks in advance!

$\endgroup$
3
$\begingroup$

It's not a size question, as the same size question could be asked for the usual Yoneda lemma.

To understand the exercise, it is important to recall what the duality principle tells us in category theory: since for any category $C$, $C^{op}$ is also a category, any theorem of the form "For any category $C, P(C)$" where $P(C)$ is a property that a category may or may not have, has a dual theorem "For any category $C, P^{op}(C)$", where $P^{op}$ is a property of a category such that $P^{op}(C) \iff P(C^{op})$

Now the point is that we may dualize when we talk about arbitrary categories. Here, the Yoneda lemma is about functors into $\mathbf{Set}$, and so natural transformations between such functors are families of arrows of $\mathbf{Set}$.

Now, $\mathbf{Set}$ is a fixed category so you may not formally dualize its arrows, or better yet : you can, but you don't get the same category.

In other words if you try to dualize the Yoneda lemma, which you express as property $P$ (a property which mentions $\mathbf{Set}$), then property $P^{op}$ will be true of any category, but $P^{op}$ will be about $\mathbf{Set}^{op}$, not $\mathbf{Set}$

It's similar to something in basic calculus: if a differentiable function $f$ has $\forall x, f(x) = x^2$, then you can deduce that $f'(0) = 0$; however, it's not the case that if $f(0 ) = 0^2$, $f'(0) = 0$

$\endgroup$
  • $\begingroup$ (+1) Thank you for the great insight! $\endgroup$ – mrtaurho Oct 8 '19 at 14:51

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.