# (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).


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?

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(AB4) usually means that arbitrary coproducts exist and are exact. What you mean is existence of kernels and cokernels, which is Grothendieck's axiom AB 1). // Note also that their complex $\cdots \to 0 \to \mathbb Z \to \mathbb Z \to 0 \to \cdots$ is isomorphic to the zero complex in $\mathcal{K(A)}$. –  Martin May 31 '13 at 16:22
@Martin Lapsus. Thanks! –  Andrea Gagna May 31 '13 at 17:43

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.
I cannot see why, if the kernel of the homopoty class of $f$ exists in $\mathcal{K}(\mathbb Z\text{-}\mathbf{Mod})$, then it should be the homotopy class of $\text{ker}(f)$. –  Andrea Gagna May 31 '13 at 20:15
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 $\ker f$...
The kernel of $f$ exists (and hence it is unique) by the lemma in the OP since the codomain is (isomorphic to) zero. –  Martin Jun 1 '13 at 9:44