Kernel of a morphism between two chain complexes concentrated in degree 0 in the homotopy category

Question

Consider in the homotopy category of complexes of abelian groups $K(Ab)$ the following morphism: $$\begin{array}{rccccccccc} \mathbb{Z}[0]:&\dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\mathbb{Z}& \overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\dots\\&&&\downarrow&&\downarrow&&\downarrow\\ \frac{\mathbb{Z}}{2\mathbb{Z}}[0]:&\dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\frac{\mathbb{Z}}{2\mathbb{Z}}& \overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\dots\\\end{array}$$ where the only non-trivial vertical map is the canonical projection. I would like to prove that it has no kernel in the homotopy category.

My definition of kernel: in a category $C$ with $0-$morphisms, we say that the kernel of a morphism $\alpha:X\to Y$ is given by an object $K$ and a morphism $\kappa:K\to X$ with the following properties:

(1) $\alpha\kappa=0$

(2)[Universal property of kernel] If $K’\in C$ and $\kappa’:K’\to X$ satisfy $\alpha\kappa’=0$ there exist a unique morphism $\Gamma:K’\to K$ such that $\kappa’=\kappa\Gamma.$

My attempt: in the category of complexes of abelian groups the map has a kernel, which is given by the complex $$2\mathbb{Z}[0]:\ \ \dots\overset{}{\rightarrow}0 \overset{}{\rightarrow}2\mathbb{Z} \overset{}{\rightarrow}0 \overset{}{\rightarrow}\dots\\ $$ and the morphism $\imath[0]:2\mathbb{Z}[0]\to\mathbb{Z}[0]$ in which the only non-trivial map is the inclusion $\imath:2\mathbb{Z}\to \mathbb{Z}.$ I think that, if the kernel exists in the homotopy category, it should be given by the same complex $2\mathbb{Z}[0]$ and by the homotopy class of the map $\imath[0].$ Given this, the proof that the kernel doesn’t exist is easy. If it is true, how can I prove this fact? Or is there another simple way to prove the non-existence of the kernel?

Answer 1

Greg Stevenson really gets credit for all of this. Imitating his answer here:

Suppose that $\kappa: K \to \mathbb{Z}[0]$ is the kernel of $\mathbb{Z}[0] \to \mathbb{Z}/2[0]$. Then $\kappa$ must be a monomorphism. Any map can be extended into an exact triangle, so do that with $\kappa$: there is an exact triangle $$ W \xrightarrow{f} K \xrightarrow{\kappa} \mathbb{Z}[0] \to \Sigma W $$ where $W$ is some chain complex and $\Sigma W$ is $W$ shifted up by one. Since this is an exact triangle, $\kappa \circ f = 0$, and since $\kappa$ is a monomorphism, this means that $f$ must be the zero map. Therefore (by another of Greg Stevenson's answers here), the exact triangle $$ K \xrightarrow{\kappa} \mathbb{Z}[0] \to \Sigma W $$ must split: $\mathbb{Z}[0] \simeq K \oplus \Sigma W$. The splitting arises as follows: since the map $f$ is zero, there is a map of exact triangles $$ \begin{array}{c} K & \to & K \oplus \Sigma W & \to & \Sigma W \\ \downarrow & & \downarrow & & \downarrow \\ K & \to & \mathbb{Z}[0] & \to & \Sigma W \end{array} $$ Since the vertical maps on the sides are isomorphisms, so is the middle one (essentially by the five-lemma).

This is a contradiction, since $\mathbb{Z}[0]$ is not decomposable as a direct sum. (If it were decomposable, well, since its homology is not decomposable, then the homology of $K$ is either zero or $\kappa$ induces a homology isomorphism. You can rule out both of these cases by using the assumption that $\kappa$ is supposed to be the kernel of the original map.)

Answer 2

I talked with my teacher, and following him I try to answer to my question (without using triangulated categories).

Suppose a kernel $\phi[0]:K^\bullet\to\mathbb{Z}[0]$ exists. We can consider the following morphisms of complexes, $\gamma^\bullet,0^\bullet,$ given by $$\begin{array}{ccccccccc} \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\mathbb{Z}&\rightarrow&\frac{\mathbb{Z}}{2\mathbb{Z}} &\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\dots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\mathbb{Z}& \overset{}{\rightarrow}&0& \overset{}{\rightarrow}&0&\rightarrow&\dots \end{array}$$ and $$\begin{array}{ccccccccc} \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&0&\rightarrow&\frac{\mathbb{Z}}{2\mathbb{Z}} &\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\dots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\mathbb{Z}& \overset{}{\rightarrow}&0& \overset{}{\rightarrow}&0&\rightarrow&\dots \end{array}$$ where the non-trivial maps $\mathbb{Z}\to\frac{\mathbb{Z}}{2\mathbb{Z}}$ and $\mathbb{Z}\to\mathbb{Z}$ are, respectively, the projection and the identity. Denote $X^\bullet$ the complex in the top row. Clearly, $\phi[0]\gamma^\bullet\sim0,$ and $\phi[0]0^\bullet\sim0.$ Consider the first morphism: by the universal property of the kernel, there exists a morphism of complexes $\alpha^\bullet:X^\bullet\to K^\bullet,$ unique up to homotopy, such that $\phi[0]\alpha^\bullet\sim \gamma^\bullet:$

$$\begin{array}{ccccccccc} \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\mathbb{Z}&\rightarrow&\frac{\mathbb{Z}}{2\mathbb{Z}} &\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\dots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ \dots&\overset{}{\rightarrow}&K^{-1}& \overset{}{\rightarrow}&K^0& \overset{}{\rightarrow}&K^1& \overset{}{\rightarrow}&K^2&\rightarrow&\dots\\&&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\mathbb{Z}& \overset{}{\rightarrow}&0& \overset{}{\rightarrow}&0&\rightarrow&\dots\\ \end{array}$$ In particular, $\text{Id}_\mathbb{Z}\sim\phi\alpha^0:$ since the only homomorphism $\frac{\mathbb{Z}}{2\mathbb{Z}}\to\mathbb{Z}$ is the zero morphism, we immediately have $\text{Id}_\mathbb{Z}=\phi\alpha^0.$ Consider the elements $a=\alpha^0(1),$ $b=\alpha^1([1]).$ By the commutativity of the diagrams we immediately have $\partial_K^0(a)=b,$ and also $$\partial_K^0(2a)=2b=2\alpha^1([1])=\alpha^1([0])=0.$$ We can now prove that the morphism $0:\frac{\mathbb{Z}}{2\mathbb{Z}}[1]\to\mathbb{Z}[0]$ doesn't have a unique factorization up to homotopy, having in this way a contradiction. Consider the following morphisms of complexes, $\psi_1^\bullet,\psi_2^\bullet:$ $$\begin{array}{ccccccccc} \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&0&\rightarrow&\frac{\mathbb{Z}}{2\mathbb{Z}} &\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\dots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ \dots&\overset{}{\rightarrow}&K^{-1}& \overset{}{\rightarrow}&K^0& \overset{}{\rightarrow}&K^1& \overset{}{\rightarrow}&K^2&\rightarrow&\dots \end{array}$$ where the non-trivial map $\frac{\mathbb{Z}}{2\mathbb{Z}}\to K^1$ is $\alpha^1,$ and $$\begin{array}{ccccccccc} \dots&\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&0&\rightarrow&\frac{\mathbb{Z}}{2\mathbb{Z}} &\overset{}{\rightarrow}&0& \overset{}{\rightarrow}&\dots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow\\ \dots&\overset{}{\rightarrow}&K^{-1}& \overset{}{\rightarrow}&K^0& \overset{}{\rightarrow}&K^1& \overset{}{\rightarrow}&K^2&\rightarrow&\dots \end{array}$$ in which even the map $\frac{\mathbb{Z}}{2\mathbb{Z}}\to K^1$ is the zero morphism. Clearly, $\phi[0]\psi_1^\bullet=0=\phi[0]\psi_2^\bullet.$ To conclude, we prove that the morphisms $\psi_1^\bullet,\psi_2^\bullet$ are not homotopic. If they were homotopic, there would be a morphism $\psi:\frac{\mathbb{Z}}{2\mathbb{Z}}\to K^0$ such that $\alpha^1=\partial_K^0\psi.$ Consider the elements $c=\psi([1],$ $d=a-c.$ We have $$\partial_K^0(c)=\partial_K^0\psi([1])=\alpha^1([1])=b,$$ hence $\partial_K^0(d)=0.$ We then have $2a,d\in\text{Ker}(\partial_K^0)$, and furthermore $$2a-2d=2c=2\psi([1])=\psi([0])=0.$$ Consider the $0-$th cohomology group of the complex $K^\bullet:$ for what we proved, we have $[2a]=[2d],$ and furthermore $$2\phi(d)=\phi(2d)=2\phi(2a)=2\phi(a)=2\phi\alpha^0(1)=2\text{Id}_\mathbb{Z}(1)=2,$$ hence we have $\phi(d)=1.$ But this is impossible: in fact, $$\text{Im}(H^0(\phi[0]))\subseteq 2\mathbb{Z}.$$