[proofplan] We prove exactness of the bottom row by a diagram chase. First we show that $\gamma_1$ is injective by lifting an element of $C_1$ to $B_1$, correcting its image in $B_2$ by an element from $A_2$, and then using exactness of the top and middle rows. Next we prove $\operatorname{im}\gamma_1=\ker\gamma_2$ by lifting an element of $\ker\gamma_2$ to $B_2$, correcting its obstruction in $B_3$ by an element from $A_2$, and descending to $C_1$. Finally, surjectivity of $\gamma_2$ follows by lifting from $C_3$ to $B_3$, using surjectivity of $\beta_2$, and descending through the middle column. [/proofplan] [step:Prove that $\gamma_1$ is injective] Let $c_1 \in C_1$ satisfy $\gamma_1(c_1)=0$. Since the first column \begin{align*} 0 \to A_1 \xrightarrow{u_1} B_1 \xrightarrow{v_1} C_1 \to 0 \end{align*} is exact, $v_1$ is surjective. Choose $b_1 \in B_1$ such that $v_1(b_1)=c_1$. By commutativity of the left lower square, \begin{align*} v_2(\beta_1(b_1)) = \gamma_1(v_1(b_1)) = \gamma_1(c_1) = 0. \end{align*} Thus $\beta_1(b_1)\in \ker v_2$. Exactness of the second column gives $\ker v_2=\operatorname{im}u_2$, so there exists $a_2\in A_2$ such that \begin{align*} u_2(a_2)=\beta_1(b_1). \end{align*} Apply $\beta_2$ to this equality. Since the middle row is exact, $\beta_2\circ\beta_1=0$, and by commutativity of the upper right square, \begin{align*} u_3(\alpha_2(a_2)) = \beta_2(u_2(a_2)) = \beta_2(\beta_1(b_1)) = 0. \end{align*} Exactness of the third column implies that $u_3$ is injective, hence $\alpha_2(a_2)=0$. Exactness of the top row gives $\ker\alpha_2=\operatorname{im}\alpha_1$, so there exists $a_1\in A_1$ such that \begin{align*} \alpha_1(a_1)=a_2. \end{align*} Using commutativity of the upper left square, \begin{align*} \beta_1(b_1-u_1(a_1)) &= \beta_1(b_1)-\beta_1(u_1(a_1))\\ &= u_2(a_2)-u_2(\alpha_1(a_1))\\ &= 0. \end{align*} Since the middle row is exact at $B_1$, $\beta_1$ is injective. Therefore $b_1=u_1(a_1)$. Applying $v_1$ and using exactness of the first column, \begin{align*} c_1 = v_1(b_1) = v_1(u_1(a_1)) = 0. \end{align*} Thus $\ker\gamma_1=\{0\}$, so $\gamma_1$ is injective. [guided] We start with the definition of injectivity. Take $c_1\in C_1$ and assume $\gamma_1(c_1)=0$; we must prove $c_1=0$. The first column is exact: \begin{align*} 0 \to A_1 \xrightarrow{u_1} B_1 \xrightarrow{v_1} C_1 \to 0. \end{align*} Exactness at $C_1$ means that $v_1$ is surjective, so we may choose an element $b_1\in B_1$ with $v_1(b_1)=c_1$. Now we push $b_1$ one step to the right. The lower left square commutes, meaning $v_2\circ\beta_1=\gamma_1\circ v_1$. Therefore \begin{align*} v_2(\beta_1(b_1)) = \gamma_1(v_1(b_1)) = \gamma_1(c_1) = 0. \end{align*} So $\beta_1(b_1)$ lies in $\ker v_2$. Exactness of the second column says $\ker v_2=\operatorname{im}u_2$, and hence there is $a_2\in A_2$ such that \begin{align*} u_2(a_2)=\beta_1(b_1). \end{align*} The element $a_2$ is the correction term in the top row. To show it comes from $A_1$, we prove it lies in $\ker\alpha_2$. Apply $\beta_2$ to the equality above. Since the middle row is exact, $\beta_2\circ\beta_1=0$. Since the upper right square commutes, $\beta_2\circ u_2=u_3\circ\alpha_2$. Thus \begin{align*} u_3(\alpha_2(a_2)) = \beta_2(u_2(a_2)) = \beta_2(\beta_1(b_1)) = 0. \end{align*} The third column is exact, so $u_3$ is injective. Therefore $\alpha_2(a_2)=0$. Exactness of the top row at $A_2$ gives $\ker\alpha_2=\operatorname{im}\alpha_1$, so there exists $a_1\in A_1$ with \begin{align*} \alpha_1(a_1)=a_2. \end{align*} Now compare $b_1$ with the image $u_1(a_1)$. The upper left square commutes, so $\beta_1(u_1(a_1))=u_2(\alpha_1(a_1))=u_2(a_2)$. Hence \begin{align*} \beta_1(b_1-u_1(a_1)) &= \beta_1(b_1)-\beta_1(u_1(a_1))\\ &= u_2(a_2)-u_2(\alpha_1(a_1))\\ &= 0. \end{align*} Exactness of the middle row at $B_1$ means $\beta_1$ is injective, so $b_1-u_1(a_1)=0$. Thus $b_1=u_1(a_1)$. Finally, \begin{align*} c_1 = v_1(b_1) = v_1(u_1(a_1)) = 0, \end{align*} because exactness of the first column gives $v_1\circ u_1=0$. This proves $\gamma_1$ is injective. [/guided] [/step] [step:Identify $\ker\gamma_2$ with $\operatorname{im}\gamma_1$] First, let $c_1\in C_1$. Since $v_1$ is surjective by exactness of the first column, choose $b_1\in B_1$ with $v_1(b_1)=c_1$. Using commutativity and exactness of the middle row, \begin{align*} \gamma_2(\gamma_1(c_1)) &= \gamma_2(v_2(\beta_1(b_1)))\\ &= v_3(\beta_2(\beta_1(b_1)))\\ &= 0. \end{align*} Thus $\operatorname{im}\gamma_1\subseteq \ker\gamma_2$. Conversely, let $c_2\in C_2$ satisfy $\gamma_2(c_2)=0$. Since the second column is exact, choose $b_2\in B_2$ such that $v_2(b_2)=c_2$. By commutativity of the lower right square, \begin{align*} v_3(\beta_2(b_2)) = \gamma_2(v_2(b_2)) = \gamma_2(c_2) = 0. \end{align*} Hence $\beta_2(b_2)\in \ker v_3=\operatorname{im}u_3$, so there exists $a_3\in A_3$ such that \begin{align*} u_3(a_3)=\beta_2(b_2). \end{align*} Since the top row is exact, $\alpha_2$ is surjective. Choose $a_2\in A_2$ such that $\alpha_2(a_2)=a_3$. Then \begin{align*} \beta_2(b_2-u_2(a_2)) &= \beta_2(b_2)-\beta_2(u_2(a_2))\\ &= u_3(a_3)-u_3(\alpha_2(a_2))\\ &= 0. \end{align*} Thus $b_2-u_2(a_2)\in \ker\beta_2$. Exactness of the middle row gives $\ker\beta_2=\operatorname{im}\beta_1$, so there exists $b_1\in B_1$ such that \begin{align*} \beta_1(b_1)=b_2-u_2(a_2). \end{align*} Set $c_1:=v_1(b_1)\in C_1$. Then, using commutativity of the lower left square and exactness of the second column, \begin{align*} \gamma_1(c_1) &= \gamma_1(v_1(b_1))\\ &= v_2(\beta_1(b_1))\\ &= v_2(b_2-u_2(a_2))\\ &= c_2. \end{align*} Therefore $c_2\in \operatorname{im}\gamma_1$, and hence $\ker\gamma_2\subseteq \operatorname{im}\gamma_1$. Combining the two inclusions gives \begin{align*} \ker\gamma_2=\operatorname{im}\gamma_1. \end{align*} [guided] We must prove two inclusions. First take an arbitrary element of $\operatorname{im}\gamma_1$, say $\gamma_1(c_1)$ with $c_1\in C_1$. Exactness of the first column gives surjectivity of $v_1$, so choose $b_1\in B_1$ such that $v_1(b_1)=c_1$. The commutative square on the lower left gives $\gamma_1(v_1(b_1))=v_2(\beta_1(b_1))$. The commutative square on the lower right gives $\gamma_2(v_2(x))=v_3(\beta_2(x))$ for every $x\in B_2$. Therefore \begin{align*} \gamma_2(\gamma_1(c_1)) &= \gamma_2(v_2(\beta_1(b_1)))\\ &= v_3(\beta_2(\beta_1(b_1)))\\ &= 0, \end{align*} because exactness of the middle row gives $\beta_2\circ\beta_1=0$. This proves $\operatorname{im}\gamma_1\subseteq\ker\gamma_2$. For the reverse inclusion, take $c_2\in C_2$ with $\gamma_2(c_2)=0$. Exactness of the second column gives surjectivity of $v_2$, so choose $b_2\in B_2$ satisfying $v_2(b_2)=c_2$. Push $b_2$ to $B_3$. By commutativity of the lower right square, \begin{align*} v_3(\beta_2(b_2)) = \gamma_2(v_2(b_2)) = \gamma_2(c_2) = 0. \end{align*} Thus $\beta_2(b_2)$ lies in $\ker v_3$. Exactness of the third column identifies this kernel with $\operatorname{im}u_3$, so there exists $a_3\in A_3$ such that \begin{align*} u_3(a_3)=\beta_2(b_2). \end{align*} The obstruction to $b_2$ being in the image of $\beta_1$ is precisely its image $\beta_2(b_2)$ in $B_3$. We remove that obstruction using the top row. Since the top row is exact at $A_3$, the map $\alpha_2:A_2\to A_3$ is surjective. Choose $a_2\in A_2$ with $\alpha_2(a_2)=a_3$. Then commutativity of the upper right square gives $\beta_2(u_2(a_2))=u_3(\alpha_2(a_2))=u_3(a_3)$. Hence \begin{align*} \beta_2(b_2-u_2(a_2)) &= \beta_2(b_2)-\beta_2(u_2(a_2))\\ &= u_3(a_3)-u_3(\alpha_2(a_2))\\ &= 0. \end{align*} So $b_2-u_2(a_2)\in\ker\beta_2$. Exactness of the middle row gives $\ker\beta_2=\operatorname{im}\beta_1$, and therefore there exists $b_1\in B_1$ such that \begin{align*} \beta_1(b_1)=b_2-u_2(a_2). \end{align*} Now descend this $b_1$ to $C_1$ by defining $c_1:=v_1(b_1)\in C_1$. We compute its image under $\gamma_1$. By commutativity of the lower left square, \begin{align*} \gamma_1(c_1) &= \gamma_1(v_1(b_1))\\ &= v_2(\beta_1(b_1))\\ &= v_2(b_2-u_2(a_2))\\ &= v_2(b_2)-v_2(u_2(a_2)). \end{align*} Exactness of the second column gives $v_2\circ u_2=0$, and $v_2(b_2)=c_2$ by construction. Therefore \begin{align*} \gamma_1(c_1)=c_2. \end{align*} Thus every element of $\ker\gamma_2$ lies in $\operatorname{im}\gamma_1$. Combining both inclusions proves $\ker\gamma_2=\operatorname{im}\gamma_1$. [/guided] [/step] [step:Prove that $\gamma_2$ is surjective] Let $c_3\in C_3$. Since the third column is exact, $v_3$ is surjective, so choose $b_3\in B_3$ such that $v_3(b_3)=c_3$. Since the middle row is exact, $\beta_2$ is surjective, so choose $b_2\in B_2$ such that $\beta_2(b_2)=b_3$. Set $c_2:=v_2(b_2)\in C_2$. By commutativity of the lower right square, \begin{align*} \gamma_2(c_2) = \gamma_2(v_2(b_2)) = v_3(\beta_2(b_2)) = v_3(b_3) = c_3. \end{align*} Therefore $\gamma_2$ is surjective. [/step] [step:Conclude exactness of the third row and state the dual form] The first step proves that $\gamma_1$ is injective. The second step proves $\operatorname{im}\gamma_1=\ker\gamma_2$. The third step proves that $\gamma_2$ is surjective. Therefore \begin{align*} 0 \to C_1 \xrightarrow{\gamma_1} C_2 \xrightarrow{\gamma_2} C_3 \to 0 \end{align*} is exact. The dual statement is obtained by applying the same argument to the transposed $3\times 3$ diagram, interchanging the roles of rows and columns throughout. [/step]