I consider a graded algebra $\{A_k\}_{k\in\mathbb{Z}}$ and suppose that the sequence
$$0\to A_{k-1}\to A_{k}\to A_{k}/A_{k-1} \to 0\qquad(*)$$
splits. I want to show that for all $j\in\mathbb{N}$, one can successively construct splittings of the sequences
$$0\to A_{k-j}\to A_{k}\to A_{k}/A_{k-j} \to 0.$$
I guess one can give proof by induction, however I have no clue where to start.
Clearly, if $p_k:A_k\to A_k/A_{k-1}$ and $q_k:A_k/A_{k-1}\to A_k$ and we know that $(*)$ splits, we have
$$A_k = ker(p_k) \oplus im(q_k) \cong A_{k-1} \oplus q_k(p_k(A_k))$$
and, similarly,
$$A_{k-1} \cong A_{k-2} \oplus q_{k-1}(p_{k-1}(A_{k-1})).$$
Therefore $$A_k\cong A_{k-2} \oplus q_{k-1}(p_{k-1}(A_{k-1})) \oplus q_k(p_k(A_k)).$$
I can iterate this but cannot see how to continue. Can somebody help me?
Since all sequences $0\to A_{k-1}\to A_{k}\to A_{k}/A_{k-1} \to 0$ split, for every $k$ there exist a homomorphism $r_k:A_k\to A_{k-1}$ such that $r_ku_{k-1}=id_{A_{k-1}}$ there $u_{k-1}:A_{k-1}\to A_k$ are inclusions.
I claim that composition $r_{k-j+1}r_{k-j+2}\dots r_k$ is the splitting homomorphism for the sequence $0\to A_{k-j}\to A_{k}\to A_{k}/A_{k-j} \to 0$. Indeed, in this sequence homomorphism $A_{k-j}\to A_k$ is equal to $u_{k-1}\dots u_{k-j}$, so $$r_{k-j+1}r_{k-j+2}\dots r_ku_{k-1}\dots u_{k-j}=r_{k-j+1}r_{k-j+2}\dots r_{k-1}u_{k-2}\dots u_{k-j}=\dots=id_{A_{k-j}}$$