Chain complex defined as a direct sum Let $(C_*,\delta_x)$ be a chain complex. Define $(D_*,d_*)$ by
$$D_n = C_{n-1} \oplus C_n,$$
$$d_n(a,b) = (-\delta_{n-1}(a),a+\delta_n(b))$$
for any $a \in C_{n-1}$ and $b \in C_n$.
$a)$ Show that $(D_*,d_*)$ is acyclic.
$b)$ Show that there exists an exact sequence of chain complexes
$$0 \to C_* \to D_* \to C_*[-1] \to 0$$
where $(C_*[-1],\delta_{{C_*[-1]}_*})$ is the chain complex where
$(C_*[-1])_n = C_{n-1}$ and $\delta_{{C_*[-1]}_n} = -\delta_{n-1}$.
I try to solve this problem. I have recently started studying homology so I don't know how to approach it. I couldn't come up with a solution for $(a)$ but for $(b)$ I did the following:
Consider the sequence
$$0 \to C_n \to D_n \to C[-1]_n \to 0.$$
Define $f_n \colon C_n \to D_n$ by $f_n(c) = (0,c)$ and $g_n \colon D_n \to C[-1]_n$ by $g_n(b,c) = -b$.
$f_n$ is inclusion map so it is clearly injective.
$g_n(0,c) = 0$ so $\text{Im}(f_n) \subseteq \text{Ker}(g_n)$.
The reverse inclusion follows by noting $g(b,c) = 0 \iff b = 0$.
Finally, $g_n$ is certainly subjective.
Hence, the sequence is exact for all $n \in \mathbb{Z}$.
Therefore, the sequence
$$0 \to C_* \to D_* \to C_*[-1] \to 0$$
is exact.
Because I didn't use the definition of $\delta_{{C_*[-1]}_n}$ here, I am not sure if there is a problem with the solution. It would be very helpful to see the mistakes I did here and also to gain insight for approaching (a).
 A: (a)
To show that $D_*$ is acyclic we need to show that its homology vanishes i.e $\text{im}(d_{n+1})=\ker(d_n)$ for all $n\in \mathbb{Z}$. The inclusion $\text{im}(d_{n+1})\subset \ker(d_n)$ holds by definition since $D_*$ is a chain complex. We will show the other inclusion.
Let $(a,b)\in \ker(d_n)$. This means that $0=d_n(a,b)=(-\delta_n(a),a+\delta_n(b))$. In particular, it holds that $a=-\delta_n (b)$. From this we see that $d_{n+1}(b,0)=(-\delta_n (b), b + \delta_{n+1}(0))=(a,b)$ which shows that $(a,b)\in \text{im}(d_{n+1})$.
We conclude that $D_*$ is acyclic.
(b)
The solution for (b) looks fine to me. However as pointed out in the comments, one should really show that $f_*$ and $g_*$ are chain maps i.e. that they commute with the differentials. That is, one should check that $d_n\circ f_n=f_{n-1}\circ \delta_n$ and $\delta_{{C_*[-1]}_n} \circ g_n = g_{n-1} \circ d_n$. This is where the definition of $\delta_{{C_*[-1]}_n}$ comes into play.
We see that
$$
d_n(f_n(c))=d_n(0,c)=(0,\delta_n(c))=f_n(\delta_n(c)),\text{ and}
$$
$$
\delta_{{C_*[-1]}_n}(g_n(b,c))=\delta_{{C_*[-1]}_n}(-b)=-\delta_{n-1}(-b)=\delta_{n-1}(b)=g_{n-1}(-\delta_{n-1}(b),b+\delta_n(c))=g_{n-1}(d_n(b,c)).
$$
This shows that $f_*$ and $g_*$ are chain maps as wanted.
