# 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 $C$?

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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.

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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
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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.

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