Exercise 1.1.3 in Charles Weibel’s book “An Introduction to Homological Algebra”

Solution 1:

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.