Is there an example of an non-essentially small over category $C\downarrow X$ of an essentially small category $C$?
Is an over category of an essentially small category again essentially small?
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0No 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. – 2012-02-18
1 Answers
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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0It'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. – 2012-02-13
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0@Thomas: Good point. I have replaced the proof with a simpler one. – 2012-02-14
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0Why is the isomorphism between $E$ and $E'$ compatible with the arrows $E\to X$ and $E'\to X$? – 2012-02-14
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0@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 $\mathcal{C} / X$. – 2012-02-14
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0@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 $p=p'\circ f$. – 2012-02-15
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0Not at all. One starts with an object of $\mathcal{C} / X$, namely $p : E \to X$, and an object of $\mathcal{C}$, $E'$, such that there is an isomorphism $f : E' \to E$ in $\mathcal{C}$. – 2012-02-15