Is it possible to realize the induced homomorphism easier? Consider chain complex exact sequence $0\to A_\cdot\to B_\cdot\to C_\cdot\to 0$. Suppose there is another chain complex exact sequence $0\to D_\cdot\to A_\cdot\to B_\cdot\to 0$
One can draw identity arrow between $A_\cdot$ and $A_\cdot$ and identity arrow between $B_\cdot$ and $B_\cdot$. 
$\textbf{Q:}$ Is there an obvious way to connect $H(C_\cdot)$ to $H(D_\cdot)[-1]$ where $H(-)$ denotes homology? I went through the following procedure. Pick a cycle $c'$ of $C_\cdot$. Lift to $b\in B_\cdot$. Because $d_Cc'=0$, I have $db=a\in A_\cdot$. Then consider $b-a$. Clearly $d_B(b-a)=0$ and $b-a$ is being sent to $c'\in C$. Hence, I can define a set theoretical map $c'\to b-a\in D_\cdot$. Is this even correct?
The goal is to check $H(C_\cdot)\cong H(D_\cdot)[-1]$.
 A: Here is what I figured out and I hope there is nothing wrong. The point is to show given $0\to B\to C\to D\to 0$ with $f:B\to C,g:C\to D$, there is quasi-isomorphism $Cone(g)\cong B[-1]$ where $B,C,D\in \mathcal{Ch}(A)$ where $A$ is abelian group or module category. Consider the following short exact sequences of chain complexes.
$0\to C\to Cyl(g)\to Cone(g)\to 0$
$0\to C\to Cone(f)\to B[-1]\to 0$
Clearly one can draw chain map $\phi:Cone(f)=C\oplus B[-1]\to Cyl(g)=C\oplus C[-1]\oplus D$ by $(c,0)\to (0,0,g(c))$ but this need not make first square commute. However note the map $\phi$ is composition of quasi-isomorphism $Cone(f)\to D\to Cyl(g)$. Now define the chain map $Cone(f)\to Cyl(g)$ by $\psi:(c,b)\to (c,-f(b'),0)$. It can be shown by brute force computation that there is a homotopy map between $\psi$ and $\phi$. Hence the vertical induced map $B[-1]\to Cone(g)$ is quasi-isomorphism.
A: I don't have any complete answer for it, but you can get at least a commutative diagram involving these homologies due to homology functor being natural: you have a commutative diagram involving exact sequences between chain complexes
$$\begin{array}
& 0 & \rightarrow & D & \rightarrow & A & \rightarrow & B & \rightarrow & 0
\\
 & & \downarrow & & \downarrow && \downarrow
\\
 0 & \rightarrow & A& \rightarrow & B & \rightarrow & C & \rightarrow & 0
\end{array}$$
which yields a commutative diagram between homologies and in particular you have commutative diagram
$$\begin{array}
& H_{n+1}(B) & \rightarrow & H_n(D)
\\
\downarrow & & \downarrow
\\
H_{n+1}(C) & \rightarrow & H_n(A)
\end{array}$$
