Commutative diagram with exact sequences as columns and rows Suppose that we have the following commutative diagram of groups and homomorphisms $$\newcommand\twoheaduparrow{\mathrel{\rotatebox{90}{$\twoheadrightarrow$}}}
\begin{array}
A  & A_3 & {\hookrightarrow} & A_2 &{\twoheadrightarrow} & A_1 & \  \\
  & \downarrow{} & &\downarrow{}& &\downarrow{}\\
 & B_3 & \stackrel{}{\hookrightarrow} &B_2 & \stackrel{}{\twoheadrightarrow} & B_1  &   & \\
  & \downarrow{} & &\downarrow{}& &\downarrow{}\\
 &C_3 &  &C_2 & \stackrel{}{\twoheadrightarrow} & C_1  &  & 
\end{array}$$
where all columns and rows are short exact sequences except the $3$rd row, where the map $C_2\to C_1$ is onto. (Sorry but not sure how to draw vertical arrows to denote $1$-$1$ maps and onto maps.)
Question

Is there a induced map $f:C_3\to C_2$ making the $3$rd row also a
  short exact sequence? Or at least can we prove that $f:C_3\to C_2$ is
  one-to-one?

I tried the five lemma but it does not seem to work here.
 A: Yes, there is such a map. This can be shown by diagram chase: 


*

*For $c \in C_3$ choose $b \in b_3$ s.t. $b=(B_3\to C_3)(b)$. Then $f(c) := (B_2 \to C_2)\circ (B_3 \to B_2)(b)$ is well-defined. 

*$f(c) \in \ker (C_2 \to C_1)$ follows from $(C_2 \to C_1)\circ (B_2 \to C_2) \circ (B_3 \to B_2) = (B_1 \to C_1)\circ (B_2 \to B_1) \circ (B_3 \to B_2)$ since the last composition is zero. 

*To show $\operatorname{im}(f) = \ker(C_2 \to C_1)$, let $c_2 \in \ker(C_2 \to C_1)$ and choose $b_2\in B_2$ with $c_2=(B_2 \to C_2)(b_2)$. Hence $(B_2 \to B_1)(b_2) \in \ker (B_1 \to C_1) = \operatorname{im}(A_1 \to B_1)$. Hence we find $a_2 \in A_2$ s.t. $(B_2 \to B_1)(b_2)=(A_1 \to B_1) \circ (A_2 \to A_1)(a_2) = (B_2 \to B_1) \circ (A_2 \to B_2)(a_2)$. Hence $b_2 - (A_2 \to B_2)(a_2) \in \ker(B_2 \to B_1)=\operatorname{im}(B_3 \to B_2)$. Choose $b_3 \in B_3$ with $(B_3\to B_2)(b_3)=b_2 - (A_2 \to B_2)(a_2)$ and set $c_3 := (B_3 \to C_3)(b_3)$. Then $f(c_3)=(B_2 \to C_2)(b_2)=c_2$. 

*The injectivity of $f$ can be shown in the same pattern as in 3. Hint: You have to use the injectivity of $A_1 \to B_1$ and of $B_3 \to B_2$. 


Note that $f$ also makes the diagram commutative. 
