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?

  • 1
    $\begingroup$ (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 $\mathcal{K(A)}$. $\endgroup$ – Martin May 31 '13 at 16:22
  • $\begingroup$ @Martin Lapsus. Thanks! $\endgroup$ – 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.

  • $\begingroup$ 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 $\text{ker}(f)$. $\endgroup$ – Andrea Gagna May 31 '13 at 20:15

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 $\ker f$...

  • $\begingroup$ The kernel of $f$ exists (and hence it is unique) by the lemma in the OP since the codomain is (isomorphic to) zero. $\endgroup$ – Martin Jun 1 '13 at 9:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.