# Isomorphism on commutative diagrams of abelian groups

Consider the following commutative diagram of homomorphisms of abelian groups $$\begin{array} 00&\stackrel{f_1}{\longrightarrow}&A& \stackrel{f_2}{\longrightarrow}&B& \stackrel{f_3}{\longrightarrow}&C&\stackrel{f_4}{\longrightarrow}& D &\stackrel{f_5}{\longrightarrow}&0\\ \downarrow{g_1}&&\downarrow{g_2}&&\downarrow{g_3}&&\downarrow{g_4}&&\downarrow{g_5}&&\downarrow{g_6}\\ 0&\stackrel{h_1}{\longrightarrow}&0& \stackrel{h_2}{\longrightarrow}&E& \stackrel{h_3}{\longrightarrow}&F&\stackrel{h_4}{\longrightarrow} &0 &\stackrel{h_5}{\longrightarrow}&0 \end{array}$$

Suppose the horizontal rows are exact ($\mathrm{ker}(f_{i+1})=\mathrm{Im}(f_i)$)

Suppose we know that $g_4:C\rightarrow F$ is an isomorphism.

How to deduce that $D=0$?

All what I could get is that $h_3:E\rightarrow F$ is an isomorphism and $f_4:C\rightarrow D$ is surjective.

where all maps $A \to A$ are the identity.
• @Theo Buehler : i see what you mean but if i show that $f_3$ is an isomorphism then i'm done. let me ask this question: consider the commmutative diagram of group homomorphisms $$\begin{matrix} A&\stackrel{f}{\rightarrow}&B\\ \downarrow{g}&&\downarrow{k}\\ C&\stackrel{h}{\rightarrow}&D \end{matrix}$$ if $h$ and $k$ are iso does this imply that $f$ is an iso? – palio Jun 7 '11 at 15:50
• @palio: no. The middle square above, that is, \begin{array}{ccc} 0 & \to & A \\ \downarrow & & \downarrow \\ A & \to & A \end{array} shows that this isn't true. However, if you know that the square is a pull-back and $h$ is an iso then you can conclude that $f$ is an isomorphism. By the way: in your original diagram $D$ is zero if and only if $f_3$ is surjective. – t.b. Jun 7 '11 at 15:57
• @palio: Could you please start accepting the answers that were helpful to you? This is done by clicking on the grey check mark sign on the left of the answers. There is a number 0% under your name showing that you haven't accepted any answers so far. People here will react much more favorably and helpfully if you have accepted at least some answers. This is how this site works. – t.b. Jun 7 '11 at 16:17