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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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closed as off-topic by user26857, Claude Leibovici, Solid Snake, Batominovski, user91500 Aug 9 at 12:47

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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

3 Answers 3

One should point out the obviousness of the first statement:

$\alpha$ is the first map of an injective composition, hence injective. $\gamma$ is the second map of a surjective composition, hence surjective.

The second statement can be also seen this way without using the snake lemma:

If $\alpha$ is an isomorphism, consider the diagram

$$\require{AMScd} \begin{CD} 0 @>>> A @>f>> B @>g>> C @>>> 0 \\ @. @V{\alpha}VV @V{\beta}VV @V{\gamma}VV @. \\ 0 @>>> {A'} @>{f'}>> {B'} @>{g'}>> {C'} @>>> 0 \\ @. @V{\alpha^{-1}}VV @V{\beta^{-1}} VV @. \\ 0 @>>> {A} @>{f}>> {B} @>{g}>> {C} @>>> 0 \end{CD} $$

By the universal property of the cokernel, we get a unique map $\delta: C' \to C$ making the diagram commutative. On the other hand, again by the universal property of the cokernel (forgetting the middle exact sequence for a while), the identity is the unique map making the outer diagram commutative. Since $\delta \circ \gamma$ has the same property, we deduce that $\delta \circ \gamma$ is the identitiy, hence $\gamma$ is injective.

The other direction of the equivalence is of course just the dual statement, so follows by abstract nonsense or can be proven the same way, using the universal property of the kernel (of course you have to add the $\prime$-sequence on the top of the diagram in that case to deduce the surjectivity of $\alpha$)

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I flagged your answer as inappropriate :). It looked weird at first sight, especially with the co-domains for the inverse functions. But I guess it is alright. It just isn't in the textbook way! –  Lehs Jan 31 at 10:12
Yes, i messed up the co-domains for the inverse-functions. Copy & Paste failure :( I will edit it. Then everything is alright. –  MooS Jan 31 at 10:29

Consider the diagram $\require{AMScd}$ \begin{CD} 0 @>>> A @>f>> B @>g>> C @>>> 0 \\ @. @V{\alpha}VV \#@V{\beta}V\!\wr V\# @VV{\gamma}V @. \\ 0 @>>> {A'} @>{f'}>> {B'} @>{g'}>> {C'} @>>> 0 \end{CD} Suppose $\alpha(a)=0$, then $\beta f(a)=0$, which implies that $a=0$, since $\beta f$ is mono. Hence $\alpha$ is mono.

Given $c'\in C'$. Since $g'$ is epi, there is a $b'\in B'$ such that $g'(b')=c'$. Now $\gamma(g\beta^{-1}(b'))=g'(b')=c'$, so $c'\in\operatorname{Im}\gamma$ and $\gamma$ is epi.

Suppose $\gamma$ is iso. Given $a'\in A'$. Then $\;\beta^{-1}f'(a')\in\operatorname{Ker}g=\operatorname{Im}f$, since $g'f'(a')=0$ and $\gamma$ is iso. Therefor it exists a $\;a\in A\;$ such that $\beta f(a)=f'(a')$. But $f'(a'-\alpha(a))=f'(a')-f'\alpha(a)=$ $=f'(a')-\beta f(a)=0$ and $f'$ is mono, so $\alpha(a)=a'$ and $\alpha$ is epi and therefor iso.

Suppose $\alpha$ is iso and that $\gamma(c)=0$. Since $g$ is epi there is a $\;b\in B$ such that $g(b)=c$. Now, $g'\beta(b)=0$ so $\beta(b)\in\operatorname{Ker}g'=\operatorname{Im}f'$ and there is a $\;a'\in A'$ such that $f'(a')=\beta(b)$. Finally, since $\alpha$ and $\beta$ are iso there is a $\;a\in A$ with $f(a)=b$ why $c=g(b)=gf(a)=0$ and $\gamma$ is iso.

I used this question to exercise diagram chasing, which is not optimal but that always work. Since it got a down vote I would like to point out that the chase is correct and could be used to learn about diagram chasing.

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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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