# Is an over category of an essentially small category again essentially small?

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

-
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

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.

-
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