Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Edited (Thanks to Kevin Carlson and Zhen Lin for pointing out the mistakes in my definitions.)

Assuming $C$ is a locally small category, Yoneda lemma says that for any given object $e \in \text{Ob}(C)$ and any functor $F: C \to \text{Set}$, the collection of natural isomorphisms from $\hom_C(e, {-})$ to $F$ is isomorphic to $F(e)$. This isomorphism is natural in $F$ and $e$.

When I try to formulate naturality rigorously, I encounter a problem about size. Here's my attempt.

Define $Y: C^{\text{op}} \to \text{Set}^C$ by $Y(c)(c') = \hom_C(c, c')$ where $\hom_C: C^{\text{op}} \times C \to \text{Set}$ is the $\hom$ bifunctor in $C$.

Define $G: \text{Set}^C \times C \to \text{Set}^C$ by $G(\eta, c) = \hom_{\text{Set}^C}(Y(c), \eta)$ where $\hom_{\text{Set}^C}: (\text{Set}^C)^{\text{op}} \times \text{Set}^C \to \text{Class}$ is the $\hom$ bifunctor in $\text{Set}^C$.

Define $E: \text{Set}^C \times C \to \text{Set}$ by $E(\eta, c) = \eta(c)$. This is the evaluation functor.

Yoneda lemma says that there exists a natural isomorphism from $G$ to $E$. My problem is that I do not know if the codomain of $G$ can be restricted to $\text{Set}$. ($\text{Class}$ is, strictly speaking, not a category, but probably a metacategory?)

My first attempt to fix this is to establish the isomorphism on each object of $\text{Set}^C \times C$ first. This will give smallness of $G(F, e) \cong F(e)$ for $F \in \text{Ob}(\text{Set}^C)$ and $e \in \text{Ob}(C)$. But then, I still have a problem proving local smallness of the image of $G$ on morphisms. Even though I know $G(F, e)$ and $G(F', e')$ are small, I do not know if the image of $G$ on $\hom_{\text{Set}^C \times C}((F, e), (F', e'))$ is small.

share|improve this question
Why are you using the internal hom? Use the external hom-functor $[\mathcal{C}, \mathbf{Set}]^\mathrm{op} \times [\mathcal{C}, \mathbf{Set}] \to \mathbf{Set}$. (Also, assume $\mathcal{C}$ is small if you want to avoid size issues.) –  Zhen Lin Jun 24 at 21:54
@ZhenLin Thank you. I guess my issue was trying to define the external hom-functor from the internal hom-functor. I cannot really see why the external hom-functor has $\mathbf{Set}$ as the codomain. Is this the consequence of $C$ being locally small? By the way, I don't want to assume that $C$ is small because most sources only assume local smallness. It also seems unnecessary to me, but may be it actually isn't... –  Tunococ Jun 24 at 23:53
$[\mathcal{C}, \mathbf{Set}]$ is locally small if $\mathcal{C}$ is small; otherwise $[\mathcal{C}, \mathbf{Set}$] may not be locally small (exercise). But if you really want to deal with that, use Mac Lane's $\mathbf{Ens}$ device. –  Zhen Lin Jun 25 at 7:17

1 Answer 1

up vote 2 down vote accepted

Naturality in the object argument just says that if $f:A\to A'$ is a morphism in $C$ then the map $f:\text{Nat}(\hom(A,-),F)\to\text{Nat}(\hom(A',-),F)$ sits in a commutative square with $Ff:FA\to FA'$, and similarly for the functor argument. This is really all you need.

But if you want a natural transformation between functors, it must of course be between roughly the functors you describe, but as quickly became apparent your codomains are not correct. There is indeed a hom bifunctor $(\text{Set}^C)^{op}\times \text{Set}^C\to \text{Set}^C$: we have $\hom_{\text{Set}^C}(F,H)(A)=\hom_{\text{Set}}(FA,HA)$. But this bifunctor has nothing to do with the Yoneda lemma-there are no natural transformations in sight! Rather, for the functor you called $G$ you want the assignment $(F,H)\mapsto\text{Nat}(F,H),$ which takes values in sets. Then your problem evaporates.

Your codomain for $\hom_C$ is more problematic still, because in general there's no way at all to make the hom functor take values in $C$: local smallness is exactly the condition that it take values in $\text{Set}$, whereas in general it's not defined as a functor at all.

share|improve this answer
Thank you. I see the mistake now. I will edit the question accordingly. I still have an issue with size though... –  Tunococ Jun 24 at 23:59
I'm not sure what you feel you need to check regarding size and morphisms. Once you've established there's only a set of natural transformations between $\hom(A,-)$ and $F$, then $G$ lands in $\text{Set}$ because its image on morphisms consists of functions between fixed sets, which form sets. –  Kevin Carlson Jun 25 at 0:29
Thank you again. I just had difficulties understanding what morphisms between sets of natural transformations are. Your way of thinking seems to work better. I should just view the morphisms as functions. –  Tunococ Jun 25 at 0:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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