Prove an isomorphism via abstract nonsense Suppose we are working in an abelian category and we have a commutative diagram with exact rows $$ \newcommand{\ra}[1]{\kern-1em\xrightarrow{#1}\kern-1em} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} 0 & \;\;\ra{} & A_1 & \ra{} & A_2 & \;\;\ra{} & A_3 & \;\;\ra{} & A_4 & \;\;\ra{} & 0 \\\ & & \da{\;f_1} & & \da{\;f_2} & & \da{\;f_3} & & \da{\;f_4} & & \\\ 0 & \;\;\ra{} & B_1 & \;\;\ra{} & B_2 & \;\;\ra{} & B_3 & \;\;\ra{} & B_4 & \;\ra{} & 0 \\\ \end{array} $$ with $f_1$, $f_3 $ and $ f_4$ isomorphisms. Can we then conclude (by diagram chasing maybe?) that $f_2$ is also an isomorphism?

Thanks for pointing it out, it's indeed an immediate application of the 5 lemma. As Matt said, just add a pair of zeros on the left to get the commutative diagram

$$ \
ewcommand{\ra}[1]{\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\xrightarrow{\quad#1\quad}\\!\\!\\!\\!\\!\\!\\!\\!} \
ewcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} 0 & \;\;\ra{} \;& 0 & \;\ra{} & A_1 & \;\;\ra{} & A_2 & \;\;\ra{} & A_3 & \;\;\ra{} & A_4 & \;\;\ra{} & 0 \\\ & & \da{\;f_0} & & \da{\;f_1} & & \da{\;f_2} & & \da{\;f_3} & & \da{\;f_4} & & \\\ 0 & \;\;\ra{} \;& 0 & \;\ra{} & B_1 & \;\;\ra{} & B_2 & \;\;\ra{} & B_3 & \;\;\ra{} & B_4 & \;\ra{} & 0 \\\ \end{array} $$

Now it's clear that the five lemma implies that $f_2$ is an isomorphism.