# 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?

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This follows trivially from five lemma, – xyzzyz Nov 14 '13 at 11:44
Doesn't this follow from the five lemma? – Alex Youcis Nov 14 '13 at 11:44
To add a little bit more to the last two comments - add two extra zeros on the left, and an isomorphism between the two you already drew. Then you can use the five lemma. – Matthew Pressland Nov 14 '13 at 11:45
Interesting to know that \uparrow and \downarrow can be used for \left and \right! – Lord_Farin Nov 14 '13 at 12:46

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} 0 & \;\;\ra{} \;& 0 & \;\ra{} & A_1 & \;\;\ra{} & A_2 & \;\;\ra{} & A_3 & \;\;\ra{} & A_4 & \;\;\ra{} & 0 \\ & & \da{\;f_0} & & \da{\;f_1} & & \da{\;f_2} & & \da{\;f_3} & & \da{\;f_4} & & \\ 0 & \;\;\ra{} \;& 0 & \;\ra{} & B_1 & \;\;\ra{} & B_2 & \;\;\ra{} & B_3 & \;\;\ra{} & B_4 & \;\ra{} & 0 \\ \end{array}$$
Now it's clear that the five lemma implies that $f_2$ is an isomorphism.