From Rotman's Algebraic Topology:
Consider a commutative diagram with exact rows:
$\dots \rightarrow A_n \xrightarrow {i_n} B_n \xrightarrow {p_n} C_n \xrightarrow {d_n} A_{n-1} \rightarrow \dots$
$\dots \rightarrow A_n' \xrightarrow {j_n} B_n' \xrightarrow {q_n} C_n' \xrightarrow {d_n} A_{n-1}' \rightarrow \dots$
with arrows $f_n,g_n$ and $h_n$ connecting $A_n \rightarrow A_n'$, $B_n \rightarrow B_n'$, and $C_n \rightarrow C_n'$. Let $h_n$ be an isomorphism.
Then there is an exact sequence $\dots \rightarrow A_n \xrightarrow {(i_n, f_n)} B_n \oplus A'_n \xrightarrow {g_n - j_n} B_n' \xrightarrow {d_n h^{-1}_nq_n} A_{n-1} \ \rightarrow \dots$.
In particular, I'm trying to show that $\text{ker } (g_n - j_n) \subset \text{im } ((i_n, f_n))$.
I know I have to show $(b,a') \in \text{ker } (g_n - j_n) \Rightarrow \exists a \in A_n$ such that $i_n(a) = b$ and $f_n(a) = a'$.
Since $gb = ja' \Rightarrow qgb=qja' = 0 \Rightarrow hpb = 0 \Rightarrow pb = 0 \Rightarrow b \in \text{ker $p_n$} = \text{im }i_n \Rightarrow ia=b$.
But I'm having trouble showing that $f(a) = a'$.
I figure:
$ia=b \Rightarrow gia = gb = jfa = ja' $
$\Rightarrow j(f(a) - a') = 0 \Rightarrow f(a) - a' \in \text{ker } j = \text{im } \Delta $
$\Rightarrow \Delta c' = f(a) - a' \Rightarrow \Delta h c = fa - a'$
But from here I'm having trouble finding a way to show that $fa - a' = 0$.
Anyone have any hints or suggestions?
I assume you meant $d$ when you wrote $\Delta$.
Assuming that, you have $\Delta h = f\Delta$, so you have $f(-\Delta c + a) = a'$
Moreover, $i(a-\Delta c) = ia - i\Delta c = ia = b$. So it's not quite $a$ that's the antecedent, but $a-\Delta c$