# Size of a natural transformation and the Yoneda Lemma

Appearing on the second page (under the section Digression: Size worries) of the following paper about the Yoneda Lemma:

http://www.maths.ed.ac.uk/~tl/categories/yoneda.ps

It says that $$a$$ $$priori$$ $$[\mathcal C^{op},Set](H_A,X))$$ is a class. I don't understand why this is the case.

My understanding is that $$a$$ $$priori$$ each member of $$[\mathcal C^{op},Set](H_A,X))$$ is a class.

So it seems to me that $$[\mathcal C^{op},Set](H_A,X))$$ could be a collection of proper classes.

I've looked at this How does the Yoneda lemma imply that $\mathrm{Hom}(yC,P)$ is a set?, but it hasn't helped me much.

Any help is appreciated -Thanks

So it seems to me that $[\mathcal C^{op},Set](HA,X)$ could be a collection of proper classes.
You're right. Each natural transformation $\alpha\colon H_A\to X$ is a class and proper when $Ob(\mathcal C)$ is proper. Therefore $[\mathcal C^{op},X](H_A,X)$ is apriori a conglomerate, see p 15-16 in Joy of Cats, which is just a extension of the class concept, much as in the same way class was an extension of set. Yoneda's lemma shows that this conglomerate is a small conglomerate and for all practical purposes can be considered a set.
• Okay, thanks a lot. Let me see if I have this straight then (using the notation of the PDF): If $[\mathcal C^{op},Set](H_A,X)$ is a class, then the Yoneda Lemma shows that $[\mathcal C^{op},Set](H_A,X)$ is an object in $Sets$ and finds a bijection (ie isomorphism) between $[\mathcal C^{op},Set](H_A,X)$ and $X(A)$ in $Sets$. Otherwise, the Yoneda Lemma shows there is a "bijection" (ie one-to-one correspondence) between $[\mathcal C^{op},Set](H_A,X)$ and $X(A)$, which is not necessarily a bijection (ie isomorphism) in $Set$. Is my understanding correct? – user52534 Mar 26 '14 at 4:16
• I wouldn't say $[\mathcal C^{op}, Set](H_A,X)$ is an OBJECT in $Set$, but rather it is essentially a set, which means there is a bijection between its members (which are classes) and the members (set elements) in an object in $Set$, namely $X(A)$. – Rachmaninoff Mar 26 '14 at 4:39
• It is similar to saying that the conglomerate (Dr. Leinster uses class here) $\{Group\}$ which consists of one class, the class of all groups, is essentially a set since there is a bijection $1\to \{Group\}$. You can't say $\{Group\}$ is a set, since its members are not sets. But if you are just collectivizing one thing, it is reasonable to call it essentially a set with one member. – Rachmaninoff Mar 26 '14 at 4:43
• By "class" I mean the set-theoretic notion of a class, and my understanding is: when $[\mathcal C^{op},Set](H_A,X)$ is a class, then the existence of a bijection between the set $X(A)$ and the conglomerate $[\mathcal C^{op},Set](H_A,X)$ implies that $[\mathcal C^{op},Set](H_A,X)$ is a set and therefore an object in $Set$. – user52534 Mar 26 '14 at 6:20
• Also, in the case that $[\mathcal C^{op},Set](H_A,X)$ is not a set, then why do we use the isomorphism symbol to say there is a bijection, in the Yoneda Lemma? Isomorphism seems to imply that there is a bijection in $Set$. But as we've discussed, the bijection need not be in $Set$.-Thanks – user52534 Mar 26 '14 at 6:26