Sign up ×
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.

In Borceux and Dejean's paper Cauchy completion in category theory, they conclude the smallness of $\bar{C}$, the full subcategory of $[C^{op},\mathrm{Set}]$ spanned by all the retracts of the representable functors, from the well-poweredness of $[C^{op},\mathrm{Set}]$. (They're assuming $C$ small.)

  1. It suffices to consider the retracts of the functors $C(-,r)$ with $r\in C$, since in a category $B$, if two objects $a,b\in B$ are isomorphic, then $\mathrm{Rct}(a)=\mathrm{Rct}(b)$ (where $\mathrm{Rct}(a)$ is the class of all the retracts of $a$; the same for $b$ with $\mathrm{Rct}(b)$).
  2. In a category $B$, if $a,b\in B$ and $b\in\mathrm{Rct}(a)$, if $r_b, r'_b:a\rightarrow b$ and $i_b,i'_b:b\rightarrow a$ are arrows in $B$ such that $r_bi_b=1$ and $r'_bi'_b=1$, then $i_b$ and $i'_b$ not necessarily represent the same subobject. Consider $3$ and $2$ in $\mathrm{Set}$.
  3. In a category $B$, given $a,b,c\in B$, if $b,c\in\mathrm{Rct}(a)$, and $b$ and $c$ represent the same subobject of $a$ then $b\cong c$. However, because of 2, if $b,c\in\mathrm{Rct}(a)$ and $b\cong c$, then $b$ and $c$ not necessarily represent the same subobject of $a$.

How to count the retracts of representables? Since isomorphism between retracts doesn't imply equality of subobjects, we should count them all, n'est-ce pas ?, but what should we do when we count twice retracts representing the same subobject?

Briefly, I don't see how to conclude the smallness of $\bar{C}$ from the well-poweredness of $[C^{op},\mathrm{Set}]$. How do Borceux and Dejean do that?

share|cite|improve this question
If you really want to pedantic, then what they have shown is that the Cauchy completion is essentially small. But that is good enough for most purposes. – Zhen Lin Feb 28 '14 at 22:52
You're right: $\bar{C}$ has a small skeleton... – Quique Ruiz Mar 3 '14 at 1:49

Your Answer


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

Browse other questions tagged or ask your own question.