# Injective Cogenerator in Abelian Category

I am trying to understand the proof of the Freyd-Mitchell Embedding Theorem and got stuck on the following detail. If $\mathcal{A}$ if a left-complete Abelian category with a generator, such that every object in $\mathcal{A}$ may be embedded in an injective object, then $\mathcal{A}$ has an injective cogenerator.

The proof is on page 70 of Freyd's book on Abelian categories and goes like so:

Let $G$ be a generator for $\mathcal{A}$, and let $P$ be the product of all the quotient objects of $G$. Let $P\to E$ be a monomorphism with $E$ injective. Then $E$ is an injective cogenerator. To prove it, let $A\to B$ be a non-zero map. Since $G$ is a generator there exists a map $G\to A$ such that $G\to A\to B\neq 0$. Let $I\to B$ the image of $G\to A\to B$, and $I\to P\to E$ be a monomorphism (this is the part I don't understand). Since $E$ in injective there exists a map $B\to E$ such that $I\to B\to E=I\to P\to E$. Now $A\to B\to E\neq 0$ because $G\to A\to B\to E=G\to A\to I\to B\to E\neq 0$.

I don't understand the choice of $P$ in the first place, I don't see where it comes into the proof. I can only assume it is used to allow the choice of the monomorphism $I\to P\to E$, but I'm not sure why. Can anyone clear this up for me?

$I\to B$ is defined as the image of a map $G\to B$; in an abelian category, this can be obtained as the factorization of $G\to B$ through the cokernel of its kernel. So $I$ is a quotient of $G$, and thus it must be a subobject of $P$, since $P$ is the product of all quotients of $G$. Then $I\to P\to E$ is a mono, since it is the composition of two monos.
This explains the choice of $P$: you need an object which has all the quotients of $G$ as subobjects. Thus the simplest choice is to take the product.
• Thank you, that makes it very clear. Can I trouble you on a couple more things; we need to know the family of quotient of objects of $G$ is a set, Freyd proves the family of subobjects of any object in an Abelian category with a generator is a set and then claims later this implies the family of quotient objects of the generator is a set, but I don't see how this follows. Mar 31 '18 at 16:22
• Also, in proving the family of subobjects is a set he says a subobject $A'\to A$ is distinguished by $(G, A')\subset (G, A)$, but I don't see this either as $(G, A')=(G, A'')$ doesn't imply $A'=A''$, maybe they are the same as subobjects though, although I'm not sure why. Mar 31 '18 at 16:26
• Ah yes, fantastic! About the second comment, $(G, -)$ being faithful implies any two distinct monomorphisms between $A'$ and $A$ are sent to separate monomorphisms (which implies the class of monomorphisms $A'\to A$ is a set, though that is already known), but when the domain is different I really don't see what can be said. Mar 31 '18 at 22:47
• @MadChickenMan If $(G,A')=(G,A'')$ as subobjects of $(G,A)$, then an arrow $G\to A$ factors through $A'$ iff it factors through $A''$. In particular, for any arrow $G\to A'$, the composition $G\to A'\to A$ factors through $A''$, and thus $G\to A'\to A\to A/A''$ must be $0$. Then $G$ being a generator implies that $A'\to A\to A/A''$ is $0$, and thus that $A'\subset A''$ (as subobjects of $A$). In the same way, we find $A''\subset A'$, and thus $A'=A''$. Apr 1 '18 at 17:35