I was reading the book Triangulated Categories by Thorsten Holm, Peter Jørgensen, Raphaël Rouquier. I found in the book the example below.

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The author wants prove that $K(\mathbf{Ab})$ is not abelian and in order to do this he takes a zero-arrow $f$ and he claims that $f$ has not kernel. Is this example wrong? If I take the zero arrow $0\colon A\longrightarrow B$ in any additive category, then is it true that it has kernel given by the identity map of $A$?

  • $\begingroup$ I'm pretty sure this example is wrong, since there is no contradiction when $r = 1$. See this question. $\endgroup$ – Takumi Murayama Nov 16 '17 at 3:49
  • $\begingroup$ Yes, the zero arrow in any additive category has a kernel, by the usual definition of kernel. So it looks like the book is talking a load of nonsense. Unless perhaps they are using some unconventional definition of kernel (but why on earth should they do that?). $\endgroup$ – Angina Seng Nov 16 '17 at 3:51
  • $\begingroup$ @LordSharktheUnknown Nope. The definition is the standard one. $\endgroup$ – Vincenzo Zaccaro Nov 16 '17 at 3:54

Yes, this example is just wrong. You are correct that the identity map on $A$ is always a kernel of the zero map $A\to B$. Their argument seems to assume that $r\mathbb{Z}$ cannot be all of $\mathbb{Z}$, but this is of course incorrect if $r=\pm 1$, which is exactly what happens for the identity map.


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