Question about an exact sequence 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.
 A: 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 $Y\to Z$.
A: 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). 
