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How can I prove that if

$$0\longrightarrow\mathrm{Hom}(M,A)\xrightarrow{\;\;i_*\;\;}\mathrm{Hom}(M,B)\xrightarrow{\;\;j_*\;\;}\mathrm{Hom}(M,C)$$ is left exact, then $$0\longrightarrow A\xrightarrow{\;\;i\;\;} B\xrightarrow{\;\;j\;\;} C$$ is left exact. I have seen proofs showing that if the second chain is left exact, then the first chain is left exact, but what about proving the converse without depending on the projective module concept.

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Your notation is confusing, can your please state a little more clearly what you are asking? What are $i,j$? Are they functors? –  user38268 Dec 28 '12 at 1:39
    
Use the Yoneda lemma. –  Zhen Lin Dec 28 '12 at 1:42
    
Closely related math.stackexchange.com/questions/235372/… –  user26857 Dec 28 '12 at 19:38

2 Answers 2

Suppose the sequence $$\tag{$*$} 0 \longrightarrow \textrm{Hom}(M, A) \longrightarrow \textrm{Hom}(M, B) \longrightarrow \textrm{Hom}(M, C)$$ is exact for all $M$. First, we must show that the sequence $$\tag{$\dagger$} 0 \longrightarrow A \longrightarrow B \longrightarrow C$$ is a (co)chain complex. So, take $M = A$ and consider $\textrm{id}_A$. We know $j_* \circ i_* = 0$, so in particular $j_* (i_* (\textrm{id}_A)) = j \circ i \circ \textrm{id}_A = 0$, so $j \circ i = 0$.

Next, we must show that $i$ is the kernel of $j$. Let $M$ be arbitrary and suppose $g : M \to B$ is a homomorphism such that $j \circ g = 0$. Then $j_* (g) = 0$, so exactness of $(*)$ means there is a unique homomorphism $f : M \to A$ such that $i_* (f) = g$, so $g = i \circ f$ for a unique $f : M \to A$. Hence $i : A \to B$ is indeed the kernel of $j : B \to C$, and therefore $(\dagger)$ is exact.

This argument works in any additive category. If we assume that we are in an abelian category then we only need to know that $(*)$ is exact for $M = A$, $M = \operatorname{Ker} i$, and $M = \operatorname{Ker} j$. Indeed, first suppose $k : M \to A$ is the kernel of $i : A \to B$. Then, $i \circ k = 0$; but there by exactness of $(*)$ that means $k = 0$, so $i$ is monic. Now suppose $g : M \to B$ is the kernel of $j : B \to C$. Then, $g \circ j = 0$, so by exactness of $(*)$ there is a unique $f : M \to A$ such that $i \circ f = g$; on the other hand, $j \circ i = 0$, so there is a unique $h : A \to M$ such that $g \circ h = i$; but $g$ is monic, so $g \circ h \circ f = g$ implies $h \circ f = \textrm{id}_M$, and $i$ is monic as well, so $i \circ f \circ h = i$ implies $f \circ h = \textrm{id}_A$, and therefore $(\dagger)$ is exact.

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A couple of questions: How do you show $i$ is injective? and how does the existence of an unique $f:M\to A$ implies $kerj \subseteq Im i$? I think the same trick of considering a specific module in each case works (M = A for the injectivity and M = B for the second), but I don't feel comfortable doing it. Thanks. –  hjhjhj57 Aug 28 at 8:30
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In a category of modules, monomorphisms = injective homomorphisms. –  Zhen Lin Aug 28 at 9:13
    
Great, that accounts for the injectivity. Could you elaborate a little bit more about the second question? –  hjhjhj57 Aug 29 at 5:03
    
It suffices to check $j \circ i = 0$. –  Zhen Lin Aug 29 at 7:20
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That is in fact true. Take $M$ to be the free module on one generator. But my proof is more general and applies to all additive categories. –  Zhen Lin Aug 30 at 7:07

The result is false without conditions on $M$. For instance if the ring in question is $\mathbb{Z}$ and $M$ is $\mathbb{Z}/2$, then the first sequence is exact for $A = \mathbb{Q}$, $B = C = \mathbb{Z}$ and the zero maps, but the second sequence isn't.

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However, if we know that the hypothesis holds for all $M$, then the claim is true. –  Zhen Lin Dec 28 '12 at 5:04
    
This is a Hom functor and the module M is being fixed, my question is not about the validity of the statement , it's rather how to prove it , I have tried to prove it but I supposed that for any mo –  JmD Dec 28 '12 at 5:22

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