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The title of my question is Proposition 3.35 in Freyd's Abelian categories. The proof says

If $G$ is a generator and $A$ is any object, then a subobject $A' \longrightarrow A$ is distinguished by the subset $(G,A') \subseteq (G, A)$.

What does "distinguished by" mean and why does this prove the proposition?

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up vote 2 down vote accepted

By "generator" Freyd surely means "strong generator". Anyway, let $A' \to A$ be a monomorphism, and consider the map $\textrm{Hom}(G, A') \to \textrm{Hom}(G, A)$. This is certainly injective, by the definition of monomorphism, so we may identify $\textrm{Hom}(G, A')$ with a subset of $\textrm{Hom}(G, A)$.

Suppose $A'' \to A'$ is another monomorphism. Then we get a chain of subsets $$\textrm{Hom}(G, A'') \subseteq \textrm{Hom}(G, A') \subseteq \textrm{Hom}(G, A)$$ and the definition of strong generator implies that $\textrm{Hom}(G, A'') = \textrm{Hom}(G, A')$ if and only if $A'' \to A$ and $A' \to A$ are isomorphic as subobjects of $A$. But what if $A' \to A$ and $A'' \to A$ are not comparable? Well, then take their pullback (which exists in any abelian category) and then use that as a bridge to compare $A' \to A$ and $A'' \to A$. The conclusion is, $\textrm{Hom}(G, A') = \textrm{Hom}(G, A'')$ as subsets of $\textrm{Hom}(G, A)$ if and only if $A' \to A$ and $A'' \to A$ are isomorphic as subobjects of $A$.

In particular, since $\textrm{Hom}(G, A)$ only has a set of subsets, $A$ only has a set of (isomorphism classes of) subobjects.

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Is it true that a Grothendieck category always has a strong generator? I'm just asking because I am sure I read somewhere that a Grothendieck category is always well-powered – Paul Slevin Jan 14 '13 at 17:17
Middle of this page:… – Paul Slevin Jan 14 '13 at 17:32
One definition of Grothendieck category asks that there be a generator. Since Grothendieck categories are locally small and cocomplete, by the remarks here, any generator is automatically a strong generator. – Zhen Lin Jan 14 '13 at 18:00

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