(The Five Lemma). Let $\require{AMScd}$ \begin{CD} A_1 @>{f_1}>> A_2 @>{f_2}>> A_3 @>{f_3}>> A_4 @>{f_4}>> A_5 \\ @V{\alpha_1}VV @V{\alpha_2}VV @V{\alpha_3}VV @V{\alpha_4}VV @V{\alpha_5}VV \\ B_1 @>{g_1}>> B_2 @>{g_2}>> B_3 @>{g_3}>> B_4 @>{g_4}>> B_5 \end{CD}
be a commutative diagram of R-modules and R-module homomorphisms, with exact rows. Prove that:
(a) $\alpha_1$ an epimorphism and $\alpha_2, \alpha_4$ monomorphisms$\implies$$\alpha_3$ is a monomorphism;
(b) $\alpha_5$ a monomorphism and $\alpha_2,\alpha_4$ epimorphisms$\implies$$\alpha_3$ is an epimorphism.
This youtube video is solution of the five lemma using diagram chasing. Short five lemma and five lemma looks very similar. I was wondering if we could prove this exercise using short five lemma. In proof of short five lemma, we make use of $g_2$ is monomorphism, which is not given in hypothesis of five lemma. We are given $\alpha_1$ is epimorphism. I want to show $g_2$ is monomorphism.
Since $\alpha_1$ is surjective, we have $\text{Im} (g_1\alpha_1)=\text{Im}(g_1)$. Since $\alpha_2$ is injective, we have $\text{ker}(\alpha_2f_1)=\text{ker}(f_1)$. By exactness at $B_2$, $\text{Im}(g_1)=\text{ker}(g_2)$. By commutativity, $\alpha_2f_1=g_1\alpha_1$. But I can’t show $\text{ker}(g_2)=\{0\}$.