Let $C(A)$ be the abelian category of complexes over the abelian category $A$. Is it true that the category of bounded above complexes with bounded homology $C^{-,b}(A)$ is abelian? Actualy I don't belive that this is true, but I saw this result in a book.
Thanks!
This is not true: $C^{-,b}(A)$ does not have kernels (unless $A$ is trivial). Let $k$ be a nonzero object of $A$ and consider the following diagram: $$\require{AMScd} \begin{CD} 0 @>>> 0 @>>> k\\ @VVV @VVV @V{1}VV\\ 0 @>>> k @>{1}>> k\\ @VVV @V{1}VV @VVV\\ k @>{1}>> k @>>> 0 \end{CD}$$
Taking the rows as three chain complexes $X_0$, $Y_0$, and $Z_0$, this diagram gives an exact sequence $0\to X_0\to Y_0\to Z_0$ of chain complexes. Now take an infinite direct sum of downward shifted copies of this exact sequence, to get another exact sequence of chain complexes $0\to X\to Y\to Z$. Then $Y$ and $Z$ will both be in $C^{-,b}(A)$, since they are bounded above and have no homology. However, the kernel $X$ of the morphism $Y\to Z$ is not an object of $C^{-,b}(A)$, since it has unbounded homology. You can then check that in fact there is no object of $C^{-,b}(A)$ which satisfies the required universal property to be a kernel of the morphism $Y\to Z$ (sketch: every summand of $X$ is an object of $C^{-,b}(A)$, so for an object to have the universal property all of these summands would have to be contained in it).