1
$\begingroup$

Hello,

I would be glad, if someone could answer a question concerning the following:

http://www.math.uni-bonn.de/people/schroer/preprints/repdim.pdf

On page 5 they show (3)=>(1). The last step is not clear to me.

I wanted to ask, why (and under which general conditions) the Hom-functor can be omitted.

Thanks.

$\endgroup$
2
  • 2
    $\begingroup$ Did you ask the author? $\endgroup$ Dec 15, 2012 at 23:47
  • 1
    $\begingroup$ The implication holds, because $M$ is a generator of $A{−}mod$. $\endgroup$ Dec 16, 2012 at 0:00

2 Answers 2

4
$\begingroup$

For any abelian category $\mathcal{A}$, an object $M\in\mathcal{A}$ is called a generator if, given an injective nonsurjective morphism $U\to V$ in $\mathcal{A}$, there always exists a morphism $M\to V$ that cannot be factorized as $M\to U\to V$.

In particular, an object $M$ in the abelian category of left $A$-modules is a generator provided that the left $A$-module $A$ is a direct summand of a direct sum of copies of $M$. The converse is not true, however (there are also generators of other types sometimes).

Given a generator $M$ of an abelian category $\mathcal{A}$ and a sequence $X\to Y\to Z$ in $\mathcal{A}$, exactness of the sequence of abelian groups $Hom_{\mathcal A}(M,X)\to Hom_{\mathcal A}(M,Y) \to Hom_{\mathcal{A}}(M,Z)$ at the middle term $Hom_{\mathcal A}(M,Y)$ implies exactness of the sequence $X\to Y\to Z$ at the middle term $Y$.

One can easily prove this claim by considering the injective morphism from the image $U$ of the morphism $X\to Y$ to the kernel $V$ of the morphism $Y\to Z$.

$\endgroup$
2
  • $\begingroup$ I always read «provided» as «only if» and get immediately confused :( $\endgroup$ Dec 16, 2012 at 3:12
  • 1
    $\begingroup$ Actually, I've made a mistake: the converse statement is true. A left $A$-module $M$ is a generator of the abelian category of left $A$-modules if and only if the left $A$-module $A$ is a direct summand of a direct sum of copies of $M$. Indeed, $M$ being a generator means that there is a surjective $A$-module map onto any $A$-module $Q$ from a direct sum of copies of $M$. When the $A$-module $Q$ is projective, it follows that $Q$ is a direct summand of a direct sum of copies of $M$. $\endgroup$ Dec 16, 2012 at 12:38
2
$\begingroup$

This is a general property of generators in Grothendieck categories (see 1.2 in this paper).

In the category of $A$-modules one can be more explicit: Let $S = \text{End}_A(M)$. Then the Hom's can be eliminated by tensoring with $M \otimes_S -$, since $$M\otimes_S \text{Hom}_A(M,N) \cong N$$ is a natrual isomorphism for each $A$-module $N$ (see 2.4c) of the linked paper).

$\endgroup$
2
  • $\begingroup$ Then one also needs to know that $M$ is a flat $S$-module (2.3 of the same paper). $\endgroup$ Dec 16, 2012 at 0:55
  • $\begingroup$ Yes, of course, see 2.1 c) ii). $\endgroup$
    – Ralph
    Dec 16, 2012 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.