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Let $A$ be a ring and let $P$ be a projective $A$-module. Then, the exactness of the sequence:

$$0\longrightarrow M_1 \overset{f}{\longrightarrow}M_2\overset{g}{\longrightarrow}M_3\longrightarrow 0 \tag{1}$$

implies the exactness of the induced $\mathbb{Z}$-module sequence:

$$0\longrightarrow \text{Hom}_A(P,\,M_1) \overset{f_\bullet}{\longrightarrow} \text{Hom}_A(P,\,M_2)\overset{g_\bullet}{\longrightarrow} \text{Hom}_A(P,\,M_3)\longrightarrow 0 \tag{2}$$

Does the exactness of (2) imply that $g$ is an epimorphism?

Let $x\in M_3$. Suppose that there exists some morphism $\varphi\in\text{Hom}_A(P,\,M_3)$ such that $x\in\text{im}(\varphi)$; then, since $g_\bullet$ is an epimorphism, there exists some $\psi\in\text{Hom}_A(P,\,M_2)$ such that $\varphi=g_\bullet(\psi)=g\circ\psi$, so $x\in\text{im}(g)$. So it suffices to show that, for every $x\in M_3$ there exists some morphism in $\text{Hom}_A(P,\,M_3)$ such that $x$ is in its image. If $P$ is free this is obviously true; is it true if $P$ isn't?

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I couldn't think of a suitable title for this, feel free to edit it. –  Fernando Martin Nov 17 '12 at 18:28
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up vote 4 down vote accepted

Well, take $P=0$, and you have a counterexample

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Is there any non-trivial counterexample? –  Fernando Martin Nov 17 '12 at 18:37
    
A slightly less trivial example is $P_1 \oplus 0$ over the ring $A_1 \times A_2$, where $P_1$ is a projective $A_1$-module... –  Zhen Lin Nov 17 '12 at 18:51
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