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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.

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

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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 $Y\to Z$.

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  • $\begingroup$ I always read «provided» as «only if» and get immediately confused :( $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 16, 2012 at 3:12
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    $\begingroup$ 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 $M$. $\endgroup$
    – Leonid Positselski
    Commented Dec 16, 2012 at 12:38
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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).

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

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