# Exact sequences of Modules

For $A$-modules and homomorphisms $0\to M′\stackrel{u}{\to}M\stackrel{v}{\to}M′′\to 0$ is exact iff for all A-modules N, the sequence

$0 \to Hom(M′′,N)\stackrel{\bar{v}}{\to} Hom(M,N)\stackrel{\bar{u}}{\to} Hom(M′,N)$ is exact.

How $\bar{u}$ and $\bar{v}$ are defined ? How do I prove that the sequence is exact at Hom(M′′,N) and Hom(M′,N)?

thanks for any help or hints

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$\bar v$ maps a $\phi:M''\to N$ to the composition $v\cdot \phi:M\to N$. – Berci Nov 10 '12 at 21:30
thats wrong it should map it to ϕ . v – user48931 Nov 12 '12 at 1:31
The claim is wrong. Correct and a trivial consequence of the universal properties is the following: $M' \to M \to M'' \to 0$ is exact iff for all $N$ we have that $0 \to \hom(M'',N) \to \hom(M,N) \to \hom(M',N)$ is exact. – Martin Brandenburg Nov 13 '12 at 13:27
@navigetor23: Unless I am mistaken, this is asking about the sequence at a different point than the other ($\mathrm{Hom}(M,N)$ instead of $\mathrm{Hom}(M'',N)$ and $\mathrm{Hom}(M',N)$). – robjohn Nov 13 '12 at 18:54

Writing composition from left to right, $\bar v$ maps $\phi\mapsto v\cdot \phi$, similarly for $\bar u$. For exactness, you need to show that $\ker\bar u={\rm im\,}\bar v$ and that $\bar v$ is injective. For example, the injectivity follows because $v$ is epimorphism, so $v\cdot\phi= 0$ implies $\phi=0$.