(AB1) failure in $\mathcal{K(A)}$, without triangulated categories I am trying to prove that there exists an abelian category $\mathcal A$ such that its homotopy category $\mathcal{K(A)}$ is (additive but) not abelian, without passing through triangulated categories (in particular, through this famous lemma).
In this beautiful book (example 2.6, pag. 10) the Authors claim a counter-example. Precisely, they consider
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{lllllllllll}
\cdots & \ra{} & 0 & \ra{} & 0 & \ra{} & \mathbb Z & \ra{} & 0 & \ra{} & \cdots \\
 & & \da{} & & \da{} & & \da{\text{id}_{\mathbb Z}} & & \da{} & & \da{} \\
\cdots & \ra{} & 0 & \ra{} & \mathbb Z & \ra{\text{id}_{\mathbb Z}} & \mathbb Z & \ra{} & 0 & \ra{} & \cdots \\
\end{array}
$$
morphism in $\mathbf{Ch}(\mathbb Z\text{-}\mathbf{Mod})$, call it $f^\bullet \colon X^\bullet \to Y^\bullet$, and claim that it(s homotopy class $[f^\bullet]_\sim$) has no kernel in $\mathcal{K(A)}$.
Evidently, $f^\bullet$ is null-homotopy, i.e., $[f^\bullet]_\sim$ is a zero-morphism, between $X^\bullet$ and $Y^\bullet$, in $\mathcal{K(A)}$. A friend of mine pointed me out this basic result:
Lemma. Let $\mathcal A$ be an additive category. Then, for every couple of objects $A, B$, a zero-morphism between them, $0^A_B \colon A \to B$ ,always has kernel.
Proof. $\text{id}_A\colon A \to A$ is a kernel. For, consider any map $g\colon X \to A$ such that $0^A_B \circ g = 0^X_B$ (i.e., any map $X \to A$).


*

*$g\circ \text{id}_A = g$

*For every $h\colon X \to A$, $h \circ \text{id}_A = g$ iff $h=g$.


Thus, $\text{id}_A$ satysfy the universal property of kernel. $\square$
Hence, the claimed counter-example must be flawed. (Actually, using their notation, they exclude $\text{Im}(k_0) \cong \mathbb Z$).
Can you provide any (working) counter-example?
 A: Let us consider :
$$\require{AMScd}
\begin{CD}
\cdots @>>> 0 & @>>> \mathbb{Z} @>>> 0  @>>> \cdots\\
@. @VVV @V{f}VV @VVV \\
\cdots @>>> 0 @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 @>>> \cdots
\end{CD}$$
and say that these are concentrated in degree 0, and we consider the map of complexes as 
a morphism in $K(\mathcal{A})$.
Now, no kernel exists - if it had one, it has to be $$\cdots \rightarrow 0 \rightarrow 2 \mathbb{Z} \rightarrow 0 \rightarrow \cdots$$ concentrated in degree zero. Call the kernel K. 
Then, we look at the follwing map in $K(\mathcal{A})$, call it g:
$$\begin{CD}
\cdots @>>> 0 & @>>> \mathbb{Z} @>>> \mathbb{Z}/2\mathbb{Z}  @>>> 0 @>>> \cdots\\
@. @VVV @V{id}VV @VVV \\
\cdots @>>> 0 @>>> \mathbb{Z} @>>> 0 @>>> \cdots
\end{CD}$$
but this leads to something absurd - namely that $fg=0$ in $K(\mathcal{A})$.This would lead to the fact that there would be a map from
$$\begin{CD}
\cdots @>>> 0 & @>>> \mathbb{Z} @>>> \mathbb{Z}/2\mathbb{Z}  @>>> 0 @>>> \cdots\\ 
\end{CD}$$
to $K$ that is homotopic to $g$, but no such map can exist. 
A: I also struggled a lot against that very example, some time ago :)
Maybe the point is that the kernel of the map they describe is not unique, because there aro two non-homotopic complexes sharing the UMP of your $\ker f$...
