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.

Is there an example of an non-essentially small over category $C\downarrow X$ of an essentially small category $C$?

share|improve this question
No examples becuase it's clear (1) that an over category of a small category is small, (2) that a category is essentially small if and only if it is equivalent to a small category and (3) that equivalent categories have equivalent over categories. –  Omar Antolín-Camarena Feb 18 '12 at 3:48

2 Answers 2

up vote 1 down vote accepted

Edit. The previous argument had a flaw. Here is a more direct proof.

No such example exists.

Almost by definition, an essentially small category is locally small, so if $\mathcal{C}$ is essentially small, every slice category $\mathcal{C} / X$ is locally small. So we only need to show that there are only set-many isomorphism classes of objects in $\mathcal{C} / X$.

We know that $\mathcal{C}$ itself has only set-many isomorphism classes of objects: so let $\mathcal{O}$ be a set of objects of $\mathcal{C}$ such that every object of $\mathcal{C}$ is isomorphic to an object in $\mathcal{O}$. But objects in $\mathcal{C} / X$ are arrows of the form $p : E \to X$ in $\mathcal{C}$, and $E$ is isomorphic to some $E'$ in $\mathcal{O}$, so $p : E \to X$ is isomorphic to some $p' : E' \to X$ in $\mathcal{C} / X$. Thus, every object in $\mathcal{C} / X$ is isomorphic to an object in the set $$\coprod_{E \in \mathcal{O}} \mathcal{C}(E, X)$$ and therefore, $\mathcal{C} / X$ is essentially small, as claimed.

share|improve this answer
It's not at all obvious that a subcategory of an essentially small category is again essentially small. For example, if you take an non-small essentially small category and take the subcategory with all the same objects but only the identity morphisms, the new category is not small. –  Thomas Andrews Feb 13 '12 at 23:00
@Thomas: Good point. I have replaced the proof with a simpler one. –  Zhen Lin Feb 14 '12 at 0:41
Why is the isomorphism between $E$ and $E'$ compatible with the arrows $E\to X$ and $E'\to X$? –  Renzo Feb 14 '12 at 8:26
@user24915: Because we choose $p' : E' \to X$ to make it compatible. This is legitimate: if $f : E' \to E$ is the isomorphism in $\mathcal{C}$, then $p$ is isomorphic to $p'$ in $\mathcal{C} / X$. –  Zhen Lin Feb 14 '12 at 10:07
@Zhen Lin: I do not understand your comment. One starts with two objects $p:E\to X$ and $p':E'\to X$ such that $f:E\to E'$ is an isomorphism in $C$ and it is to show that $p$ and $p'$ are isomorphic in $C/X$. I see no reason why $p=p'\circ f$. –  Renzo Feb 15 '12 at 8:14

The first thing to show is that if $D$ is a small category, then for any object $Y$ in $D$, $D\downarrow Y$ is a small category. I think that is trivial.

Now, if $C$ is essentially small, there is a small category $D$ which is equivalent to $C$.

That means there is are functors $F:D\rightarrow C$ and $G:C\rightarrow D$ and natural isomorphisms $FG\cong {id}_{C}$ and $GF\cong {id}_{D}$.

I don't see any big obstruction to taking this equivalence into an equivalence between $C\downarrow X$ and $D\downarrow GX$.

Namely, define $G_{X}:C\downarrow X\rightarrow D\downarrow G(X)$ by sending $f:Y\rightarrow X$ to $G(f):G(Y)\rightarrow G(X)$.

Define $F_{X}:D\downarrow G(X)\rightarrow C\downarrow X$ by sending $f:Z\rightarrow G(X)$ to $\mu_{X}\circ F(f)$, where $\mu_{X}:FG(X)\rightarrow X$ is the natural isomoporhism between $FGX$ and $X$.

Then you just have to show that $F_X$ and $G_X$ are functors and that there are natural isomorphisms $F_XG_X\cong {id}_{C\downarrow X}$ and $G_XF_X\cong {id}_{D\downarrow G(X)}$. Unless I'm missing something, this should be fairly direct.

share|improve this answer
Whoops, meant "essentially small," but got confused in the typing. Fixed. Thanks @ZhenLin –  Thomas Andrews Feb 13 '12 at 21:44

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.