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Let $A$ and $A'$ be abelian groups. Let $f\colon A \rightarrow A'$ be a surjective homomorphism. Let $B$ be a subgroup of $A$. Let $B' = f(B)$. Let $A_0 = Ker(f)$. Let $B_0 = A_0 \cap B$. Is the following sequence exact?

$0 \rightarrow A_0/B_0 \rightarrow A/B \rightarrow A'/B' \rightarrow 0$

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Perhaps this can also be proved by snake lemma? –  Makoto Kato Jun 20 '12 at 3:59

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Since $f\colon A\to A'/B'$ is onto, and factors through $A/B$, then $A/B\to A'/B'$ is onto.

Since $A_0\to A$ is one-to-one, the kernel of $A_0\to A/B$ is $A_0\cap B=B_0$, so $A_0/B_0$ embeds into $A/B$ by mapping $a+B_0$ to $a+B$.

If $a\in A_0$, then $f(a+B) = f(a)+B'$, and since $f(a)=0$, then the composition is zero.

So the only issue is whether, if $a+B$ maps to zero, then $a+B$ is in the image of $A_0/B_0$. If $a+B$ maps to zero in $A'/B'$, then $f(a)\in B'$, so there exists $b\in B$ such that $f(b)=f(a)$. Thus, $a-b\in A_0$. Since $a+B = (a-b)+B$, and $(a-b)+B$ is the image of $a-b+B_0$, this follows.

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