# A category with a strong generator is well-powered

Reading the book of Adamek and Rosicky, Locally Presentable and Accessible categories, I found an exercise I'm not able to solve (unless under some additional hypothesis).

Precisely, my problem is to prove the following statement:

Any (locally small) category with a strong generator is well-powered.

I recall my definitions: a set of objects $\mathcal G$ of a category $\mathcal C$ is said to be a strong generator if it is a generator and moreover for each object $K \in \text{Ob}(\mathcal C)$ and any proper subobject $m \colon K' \to K$ there is an element $G \in \mathcal G$ and a morphism $g \colon G \to K$ not factoring through $m$.

A category is said to be well-powered if for each object $K$, $\text{Sub}(K)$ is a set and not a proper class (I will assume to work in NBG).

Now, I'm able to prove this fact under the addtional hypothesis that $\mathcal C$ has pullbacks, but so far I haven't been able to figure out the general proof. I sketch my proof, just in the case it can be of some help in finding out the general one.

Fix an object $K \in \text{Ob}(\mathcal C)$ and consider $S := \coprod_{G \in \mathcal G} \text{Hom}_{\mathcal C}(G,K)$. This is a set because the category is locally small and $\mathcal G$ is a set by assumption. Define $F \colon \text{Sub}(K) \to \text{Parts}(S)$ by setting $$F( m) := \{(G,f) \in S \mid f \text{ does not factor through }m\}$$ I claim that $F$ is injective (and this will conclude). In fact, let $m_1 \colon K_1 \to K$ and $m_2 \colon K_2 \to K$ be two distinct subobjects. Let $m_3 \colon K_3 \to K$ be their intersection (i.e. the pullback of $K_1 \xrightarrow{m_1} K \xleftarrow{m_2} K_2$), let $j_i \colon m_3 \to m_i$ be the two canonical projections of the pullback. Then since $m_1$ is different (as subobject) from $m_2$, either $j_1$ or $j_2$ is not an isomorphism, let's say that $j_1$. Since $\mathcal G$ is a strong generator there is a morphism $g \colon G \to K_1$ not factoring through $j_1$ (for $G \in \mathcal G$); hence $m_1 \circ g$ cannot factor through $m_2$, otherwise the UMP of the pullback would force $g$ to factor through $j_1$, contradicting our construction. Hence $m_1 \circ g \in F(m_2)$ and $m_1 \circ g \not \in F(m_1)$, i.e. $F(m_2) \ne F(m_1)$.

How to get rid of the need for pullbacks?

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Johnstone [Sketches of an elephant, Remark A1.4.17] only mentions that a locally small category with finite intersections of subobjects and a (strong) generating set is well-powered, so it's possible that you do need an extra condition. –  Zhen Lin Dec 8 '12 at 19:51
That's essentially the hypothesis I added in order to make my proof work. I thank you for this confirmation. –  Mauro Porta Dec 8 '12 at 19:59

Ok, my previous answer wasn't correct. Now, everything works.

First of all, I thank Sergio Buschi, who provided me a counter-example to the OP. Since he gave me his permission, I'll write the details. Let $\mathcal C$ be the subcategory of $\mathbf{Set}$ whose objects are all the sets. As morphisms, we include the two natural morphisms $u,v \colon \{0\} \to \{0,1\}$; in addition, for every other set $S$ we choose exactly one morphism $t_S \colon \{0\} \to S$ and we add exactly one morphism $S \to \{0,1\}$ sending every element to $0$. Then the morphisms $S \to \{0,1\}$ are monomorphisms and $\{0\}$ is a strong generator (easy check). Hence $\mathcal C$ is not well-powered.

As for my problem, I realized two things:

1. it is signalated in the Errata to the book that the exercise as stated is wrong;
2. what I really needed was that if $\mathcal C$ is a locally finitely presentable category, then it is well-powered. The argument of Adamek uses the fact that $\mathcal C$ is well-powered. Indeed, this is true in a more general form: if $\mathcal C$ is a cocomplete category endowed with a small full subcategory $\mathcal B$ with the property that every element in $\mathcal C$ is a colimit of elements in $\mathcal B$, then $\mathcal C$ is complete. And my previous argument concludes. Since this holds by definition for a locally finitely presentable category, I'm really done.

I must again thank Sergio Buschi for the terse statement above. I gave a proof by myself, but in stating it I was assuming too much hypothesis.

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