I am reading a notes of Brian Conrad on hypercovering, and in the discussion on Dold-Kan correspondence, it is claimed (I might misunderstand) that the injective objects in the category of non-negative cochain complexes ($\mathcal{CoChain}_{\ge 0}(Ab)$) are exactly those complex $I^\bullet$, such that all the $I^j$ is injective (in $Ab$), and the positive cohomology vanishes.
Here is the link of the notes: https://www.google.com/url?sa=t&source=web&rct=j&url=https://math.stanford.edu/~conrad/papers/hypercover.pdf&ved=2ahUKEwjHz8rVorP0AhVoR_EDHUeJAk8QFnoECAMQAQ&usg=AOvVaw0NKji5VcAbACZX2hMZlXek. The claim appears on page 12, in the first paragraph of the proof of Cor.2.13.
Although I can see the two conditions here are necessary, and I do not know why these are sufficient conditions. In particular, if I take an object in $Ab$, and take its injective resolution, I do not think the resolution is injective. So my question is whether there is a good characterization of injective objects in this category. Should I expect there to be some simple answer? Any comment is sincerely welcome.
Ok, I think I have found the answer myself. The answer should be that one need to add one more condition, that is, $H^0(I^\bullet)$ is also injective.
First, to see why this condition is necessary, we take $A^\bullet=A[0]$ and $B^\bullet=B[0]$ with $A\hookrightarrow B$, then $$Hom(A^\bullet, I^\bullet)=Hom(A,H^0(I^\bullet))$$ Therefore, $H^0(I^\bullet)$ should also be injective.
Now given such a complex, from the process above, we see it will not harm as to add in $H^0(I^\bullet)$ as the (-1)-th term. So without loss of generality, we simply assume that $I^\bullet$ has $H^i$ vanish for all $i\ge0$. Then we have the standard process of breaking the long exact sequence into short ones, and moreover since the sequence consists of injectives, each of these short exact sequences actually split. Then a basic diagram chasing gives the result.