I am trying to teach myself some homological algebra and I got stuck right at the start with Exercise 1.1.3 from the book “An Introduction to Homological Algebra” by Charles Weibel.
Exercise 1.1.3 (Split exact sequences of vector spaces) Choose vector spaces $\{B_n, H_n\}_{n\in\mathbb{Z}}$ over a field, and set $C_n=B_n\oplus H_n\oplus B_{n-1}$. Show that the projection-inclusions $C_n\to B_{n-1}\subset C_{n-1}$ make $\{C_n\}$ into a chain complex, and that every chain complex of vector spaces is isomorphic to a complex of this form.
My take so far
The first part is just a matter of checking that the projection-inclusions $\partial_n\colon C_n\to C_{n-1}$ satisfy $\partial_n\circ\partial_{n+1}=0$, which is obvious since $\partial_n$ basically does the following: $(x,y,z)\mapsto(z,0,0)$ and so when applied twice gives zero.
The second part is trickier for me. I started with a chain complex of vector spaces $$\cdots\to V_{n+1}\stackrel{d_{n+1}}{\to}V_n\stackrel{d_n}{\to}V_{n-1}\to\cdots$$ and looked for a map $$u_n\colon V_n\to \operatorname{im} d_{n+1}\oplus (\ker d_n/\operatorname{im} d_{n+1})\oplus \operatorname{im} d_n$$ as suggested by the notation $\{B_n, H_n\}$ in the text of the exercise.
I have come up with only one reasonable map, defined as follows. Choose a basis $\mathcal{B}_1$ for $\operatorname{im} d_{n+1}$ and extend it first to a basis $\mathcal{B}_1\sqcup\mathcal{B}_2$ for $\ker d_n$, then to a basis $\mathcal{B}_1\sqcup\mathcal{B}_2\sqcup\mathcal{B}_3$ for $V_n$. As a result, each vector $v\in V_n$ can be uniquely written as $v=v_1\oplus v_2\oplus v_3$ where $v_i$ is a linear combination of elements of $\mathcal{B}_i$, $i=1,2,3$. Finally define $$u_n(v)=v_1\oplus [v_2] \oplus d_n(v_3)$$ where $[~]$ denotes the homology class.
I checked and $u_n$ is indeed an isomorphism of vector spaces, but alas it does not seem to commute with the differentials since in general $$(u_{n-1} \circ d_n)(v) = 0 \neq d_n v_3 = (\partial_n\circ u_n)(v)\;.$$
Any help is appreciated.
Actually everything you wrote is correct, except the last equation!
Why do you say $(u_{n-1} \circ d_n)(v) = 0$? In reality, $d_n v$ is an element of $C_{n-1}$ that you need to write as a sum $v_1' + v_2' + v_3'$ where $v_1' \in \operatorname{im}d_n$ etc. But you know that such a sum is unique, and you already have the decomposition $d_n v = d_n v + 0 + 0$. Therefore $u_{n-1} d_n v = (d_n v, 0, 0)$.
On the other hand, $\partial_n u_n v = (d_n v_3, 0, 0)$. But by the identification that allowed you to define $\partial_n$, namely $C_n / \operatorname{ker} d_n = \operatorname{im} d_n$, it follows that this last element is equal to $(d_n v, 0, 0)$. Everything checks out.
PS: You perhaps know this, but working over a field was essential here. Being able to write stuff as direct sums is what allows you to perform this trick; over a general ring this result is false.