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Let a commutative diagram be given:

$$\require{AMScd} \begin{CD} 0 @>>> A @>f>> B @>g>> C @>>> 0 \\ @. @V{\alpha}VV @V{\beta}VV @V{\gamma}VV @. \\ 0 @>>> {A'} @>{f'}>> {B'} @>{g'}>> {C'} @>>> 0 \end{CD} $$

The two lines are exact sequences and $\beta$ is $R$-isomorphism

Prove that $\alpha$ is monomorphism, $\gamma$ is epimorphism. Furthermore $\alpha$ is isomorphism iff $\gamma$ is isomorphism

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Not again 5 lemma question. The proof is there math.msu.edu/~rotthaus/teaching/910/homework/solhw5.pdf. Not hard to find in google. –  user52045 Nov 26 '13 at 9:12
what have you tried?? what do you know about this particular commutative diagram –  Praphulla Koushik Nov 26 '13 at 9:12
You don't want the 5 lemma, you want the snake lemma: en.wikipedia.org/wiki/Snake_lemma –  Kevin Nov 26 '13 at 9:22
Can you explain more precise? –  user109584 Nov 26 '13 at 15:00
I have made it precise, resolving this question –  Alexander Grothendieck Dec 7 '13 at 13:02

1 Answer 1

The snake lemma gives an exact sequence $0\rightarrow\mathrm{Ker} \ \alpha\rightarrow\mathrm{Ker} \ \beta\rightarrow\mathrm{Ker} \ \gamma\rightarrow\mathrm{Coker} \ \alpha\rightarrow\mathrm{Coker} \ \beta\rightarrow\mathrm{Coker} \ \gamma\rightarrow 0$. As $\mathrm{Ker} \ \beta=\mathrm{Coker} \ \beta=0$ by hypothesis, it follows that $\mathrm{Ker} \ \alpha=0$ and $\mathrm{Coker} \ \gamma=0$ by exactness of the sequence. At last the part $0\rightarrow \mathrm{Ker} \ \gamma\rightarrow\mathrm{Coker} \ \alpha\rightarrow 0$ shows that $\mathrm{Ker} \ \gamma\cong\mathrm{Coker} \ \alpha$, whence the second statement.

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