The following is taken from Module Theory An Approach to Linear Algebra by T. S. Blyth:
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Exercise 3.10: [The 3x3 lemma] Consider the following diagram of $R$-modules and $R$-morphisms $\require{AMScd}$ \begin{CD} @. 0 @. 0 @. 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> A' @>{\smash{\alpha'}}>> A @>{\smash{\beta'}}>> A'' @>>> 0 \\ @. @VV{f'}V @VV{f}V @VV{f''}V \\ 0 @>>> B' @>{\smash{\alpha}}>> B @>{\smash{\beta}}>> B'' @>>> 0 \\ @. @VV{g'}V @VV{g}V @VV{g''}V \\ 0 @>>> C' @. C @. C'' @>>> 0 \\ @. @VVV @VVV @VVV \\ @. 0 @. 0 @. 0\end{CD}
Given that this diagram is commutative, that all three columns are exact and that the top two rows are exact, prove that there exist unique $R$-morphisms $\alpha'': C'\to C$ and $\beta'': C\to C''$ such that the resulting bottom row is exact and the complete diagram is commutative.
[Hint. Observe that $g\circ \alpha \circ f' = 0$ so that $\operatorname{ker}g' = \operatorname{im}f' \subset \operatorname{ker}g \circ \alpha$. Use Theorem 3.4 to produce $\alpha''$. Argue similarly to produce $\beta''$. Now chase!]
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I would like to know for the hint in the above exercise, how does one justify $g\circ \alpha \circ f' = 0$? I mean should it not be $g\circ f \circ \alpha' = 0$ instead? I mean $g\circ f = 0$, due to exactness at $B$.
Thank you in advance
Use commutativity of the squares in the diagram together with the exactness of the vertical sequences (and of course associativity of composition of arrows in any category): $$ g \circ \alpha \circ f' = g \circ f \circ \alpha' = 0 \circ \alpha' = 0. $$