I want to prove the following lemma only by using diagram chasing:
Nine Lemma: Let us have the following commutative diagram with exact columns and exact middle and bottom rows. So the top row is also exact.
But I'm stuck on proving that morphism $B_1 \to C_1$ is monic. I tried the following: $$c_1 \to c_2 \to 0 \in C_3 (\text{by exactness}), \text{ where } c_i \in C_i$$ So for some $a_2 \in A_2$ we can take element $(-\gamma(a_2) + a_3) \in A_3$, where $\gamma : A_2 \to A_3$, so it goes to $0 \in B_3$ and there exists such $b_2 \in B_2$ and $b_1 \in B_1$ that: $$b_1 \to b_2 \to 0 \in B_3$$ But does it matter (from the diagram commutativness) that $b_1 \to c_1$ ?
Moreover, is it true that we don't need exactness of the top row to say that $A_1 \to B_1 \to C_1 = 0$ ? How can i show it ?
I'll first label the maps as follows, for the rows we want
$$ 0 \to A_i \xrightarrow{f_i} B_i \xrightarrow{g_i} C_i \to 0, $$
and for the column I'll define
$$ 0 \to A_1 \xrightarrow{f_A} A_2 \xrightarrow{g_A} A_3 \to 0, $$
and similarly for the second and third columns
Now suppose that $c_1 \in C_1$, then there exists some $b_2 \in B_2$ such that $g_2(b_2) = f_C(c_1)$. Then
$$ g_3g_B(b_2) = g_Cg_2(b_2) = g_Cf_C(c_1) = 0, $$
and so there exists some $a_3 \in A_3$ such that $f_3(a_3) = g_B(b_2)$. But then there exists $a_2 \in A_2$ such that $a_3 = g_A(a_2)$, and then
$$ g_Bf_2(a_2) = f_3g_A(a_2) = f_3(a_3) = g_B(b_2), $$
and so $b_2 - f_2(a_2) \in \ker g_B = \operatorname{img} f_B$, and so there exists some $b_1 \in B_1$ such that $f_B(b_1) = b_2 - f_2(a_2)$. Then
$$ f_Cg_1(b_1) = g_2f_B(b_1) = g_2(b_2 - f_2(a_2)) = g_2(b_2) - g_2f_2(a_2) = g_2(b_2) = f_C(c_1), $$
and so $g_1(b_1) = c_1$ since $f_C$ is injective. Hence $B_1 \xrightarrow{g_1} C_1$ is surjective.
EDIT: To add a comment on your final question. Suppose that we are given that $g_2f_2 = 0$ and that $f_C$ is monic. Then I claim that $g_1f_1 = 0$. Well it suffices to show that $f_Cg_1f_1 = 0$, but by the commutative diagram
$$f_Cg_1f_1 = g_2f_Bf_1 = g_2f_2f_A = 0,$$
and so we're done.