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.

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


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

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
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

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.