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Given the following commutative diagram of exact sequences

$$ \begin{array} & & 0 & 0 & 0 &\\ & \downarrow & \downarrow & \downarrow &\\ 0 \rightarrow & A \stackrel{u}\rightarrow & B \stackrel{v}\rightarrow & C \rightarrow & 0\\ & \downarrow f & \downarrow g & \downarrow h &\\ 0 \rightarrow & A_1 \stackrel{u_1}\rightarrow & B_1 \stackrel{v_1}\rightarrow & C_1 & \\ & \downarrow & \downarrow g_1 & \downarrow h_1 &\\ & 0 \rightarrow & B_2 \stackrel{v_2}\rightarrow & C_2 \rightarrow & 0\\ \end{array} $$

can I conclude that also $B_1\rightarrow C_1$ is surjective??

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When diagram chasing doesn't work, then you should look for a counterexample. –  user26857 Dec 18 '12 at 14:59

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The following is a counterexample: \begin{array}[ccccccccc] \ & & 0 & & 0 & & 0 & & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & 0 & \rightarrow & 0 & \rightarrow & 0 & \rightarrow & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & 0 & \rightarrow & 0 & \rightarrow & \mathbb Z & & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0 & \rightarrow & \mathbb Z & \rightarrow & \mathbb Z & \rightarrow & 0\\ \end{array}

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