Corresponding long exact sequence to boundaries-cycle exact sequence to $0 \to Z \to C \to B(-1) \to 0$ breaks up into short exact sequences. The following is Exercise 1.3.4. from Weibel's Homological Algebra.

Let $C$ be a chain complex. Then, we have the associated exact sequence $$0 \to Z \to C \to B(-1) \to 0,$$ where $Z$ is the complex of cycles (kernels) and $B$ of boundaries (images).
Show that the corresponding long exact sequence of homology breaks up into short exact
sequences.

This is right after the discussion on the long exact sequence of homologies associated to a short exact sequence. From what I understand, we have a long exact sequence of the homologies given as
$$\cdots \to H_{n}(Z) \to H_{n}(C) \to H_{n - 1}(B) \xrightarrow{\partial} H_{n - 1}(Z) \to \cdots. \tag{1}$$
So let us compute the homologies of $Z$ and $B(-1)$. From what I see, the maps $Z_{n} \to Z_{n - 1}$ and $B_{n - 1} \to B_{n - 2}$ are simply the restriction of differentials and are simply $0$. (The former is zero since $Z$ is the kernel. The latter is zero since $B$ is contained in the kernel.)
As a result, the homologies are simply the modules themselves. Thus, $(1)$ simplifies to
$$\cdots \to Z_{n} \to H_{n}(C) \to B_{n - 1} \xrightarrow{\partial} Z_{n - 1} \to H_{n - 1}(C) \to \cdots.$$
How does the above break into short exact sequences? I have a feeling that the maps $Z_n \to H_n(C)$ are simply the quotient maps, the maps $H_{n}(C) \to B_{n - 1}$ are $0$ maps, and the connecting homomorphism $\partial$ becomes the inclusion map. Is that correct?
How does one interpret that as "breaking up into short exact sequences"?
 A: "Break up into short exact sequences" usually means splitting the groups in the chain complex so that the long exact sequence is a direct sum of short exact sequences.  As you suspected, the connecting homomorphisms are inclusions, since for $b\in B_{n-1}$, there is a $c\in C_n$ with $b=\partial c$, and then the connecting homomorphism $\partial$ applied to $b$ is defined to be $\partial b = \partial c$, which is an element of $Z_{n-1}$.
The maps $H_n(C) \to B_{n-1}$ are indeed $0$ maps, since they are given by taking the differential, but elements of $H_n(C)$ are represented by cycles.
So, the long exact sequence is direct sums of the following short exact sequences, each copy shifted over by 3
$$0 \to B_n \to Z_{n} \to H_{n}(C) \to 0$$
That is, the long exact sequence looks like this:
$$\cdots \to H_{n+1}(C)\oplus0 \to 0\oplus B_n \to Z_{n} \to H_{n}(C) \oplus 0\to 0 \oplus B_{n-1} \to Z_{n-1} \to H_{n-1}(C)\oplus 0 \to \cdots $$
This is related to a construction where you take an exact sequence
$$A_{i-2} \xrightarrow{f} A_{i-1} \to A_{i} \to A_{i+1} \xrightarrow{g} A_{i+2}$$
and then extract a short exact sequence
$$0 \to \operatorname{coker}f \to A_i \to \ker g \to 0$$
In some circumstances, $\operatorname{coker}f$ and $\ker g$ split $A_{i-1}$ and $A_{i+1}$, and you can use short exact sequences obtained in this way to break up a long exact sequence.
