I begin to read about Module Theory. After reading Five Lemma and Snake Lemma, I have this following statement.
Proposition 1. Consider the following commutative diagram:
where $\alpha,\beta,\gamma$ are isomorphisms. If the top row is an exact sequence, then the bottom row is also exact.
Proof. If $f'(x)=0$ for $x\in A',$ then there exists only $y\in A$ such that $x=\alpha(y),$ and $f'(\alpha(y))=0.$ This implies $\beta(f(y))=0,$ and so $f(y)=0$ by the property of $\beta.$ Since $f$ is injective, we have $y=0,$ and so $x=0.$ Thus $f'$ is injective.
If $c'\in C',$ then there exists $c\in C$ such that $\gamma(c)=c'.$ Since $c\in C,$ there exists $b\in B$ such that $g(b)=c.$ Then $\gamma(g(b))=c',$ implies $g'(\beta(b))=c'.$ Thus $g'$ is surjective.
By the top row is an exact sequence, we have ${\rm Im}f=\ker g.$ If $y\in {\rm Im}f',$ then $y=f'(x)$ for some $x\in A'.$ Since $x\in A',$ there exists $z\in A$ such that $x=\alpha(z),$ and so $y=f'(\alpha(z))=\beta(f(z)).$ Thus $g'(y)=g'(\alpha(f(z)))=\gamma(g(f(z)))=\gamma(0)-0,$ implies $y\in\ker g'.$ Therefore ${\rm Im}f'\subseteq\ker g'.$ If $y\in\ker g',$ then $g'(y)=0.$ Then there exists $x\in B$ such that $y=\beta(x),$ and so $0=g'(\beta(x))=\gamma(g(x)).$ This implies $g(x)=0$ i.e $x\in\ker g.$ Thus $x=f(z)$ for some $z\in A,$ and so $y=\beta(f(z))=f'(\alpha(z)).$ Therefore $\ker g'\subseteq {\rm Im}f'.$
This is my proof. Futhermore, I think about a strong result. But I can't still complete this proof. Let's see.
Proposition 2. Consider the following diagram:
where $\alpha,\beta,\gamma$ are isomorphisms. If the top row is an exact sequence, then the bottom row is also exact.
Have you ever seen this result in another books?