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.
Best Answer
This is wrong. Consider
\begin{array}{ccccccccccc} 0 & \to & 0 & \to & 0 & \to & A & \to & A & \to & 0\\ \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow\\ 0 & \to & 0 & \to & A & \to & A & \to & 0 & \to & 0 \end{array}
where all maps $A \to A$ are the identity.