Consider a $\mathbb{Z}$-graded chain complex $A^{\bullet}$, I know that a bounded below complex is one such that $A^i = 0$ for $i$ sufficiently small, and a bounded above complex is one such that $A^i = 0$ for sufficiently large $i$, every bounded above (or below) complex has a Cartan-Eilenberg resolution and every unbounded complex has a resolution too (according to Spaltenstein and other authors).

I wanted to ask:

- Does this apply to unbounded complexes that are not $\mathbb{Z}$-graded? Or finite complexes? Arbitrary chain complexes? An $\mathbb{N}$-graded complex for example? Do they too have resolutions?

I just wanted to know as I don't think I know the reason as to why boundedness is such a key thing, I know that the classical construction of resolutions for complexes works only for bounded complexes but I don't know why or where it fails for unbounded complexes.

addthe zeroes, you can construct an injective resolution for that, which also has zeroes on the left, which you can then remove. – Mariano Suárez-Alvarez♦ Feb 16 '13 at 20:51