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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?

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$I\to B$ 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.

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  • $\begingroup$ 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. $\endgroup$ – MadChickenMan Mar 31 '18 at 16:22
  • $\begingroup$ 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. $\endgroup$ – MadChickenMan Mar 31 '18 at 16:26
  • $\begingroup$ For your first comment : this is because for every object it's collections of subobjects and quotients are in bijection, since every subobject is the kernel of its cokernel and every quotient is the cokernel of its kernel. $\endgroup$ – Arnaud D. Mar 31 '18 at 21:22
  • $\begingroup$ 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. $\endgroup$ – MadChickenMan Mar 31 '18 at 22:47
  • $\begingroup$ @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 $A'=A''$. $\endgroup$ – Arnaud D. Apr 1 '18 at 17:35

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