Lemma 1.2.4 on page 39 of Neeman's book Triangulated Categories states:

Suppose we are given a candidate triangle $$X\rightarrow A\oplus Y\stackrel{\left(\begin{array}{cc}1&\alpha\\ \beta&\gamma \end{array}\right)}{\longrightarrow}A\oplus Z\rightarrow\Sigma X\ \ (*).$$ Then this candidate triangle is isomorphic to a direct sum of candidate triangles $$0\rightarrow A\rightarrow A\rightarrow0\ \ \ (**)$$ $$X\rightarrow Y\rightarrow Z\rightarrow\Sigma X\ \ (***).$$

The proof given in the book is by showing that there are maps of candidate triangles $(**)\rightarrow(*)\rightarrow(**)$ that compose to the identity map of candidate triangle $(**)$, and that candidate triangle $(***)$ can be obtained as kernel of the map $(*)\rightarrow(**)$, hence it is a direct summand.

Questions. Why is this proof valid? In particular: (1) is it true that if morphisms $T\rightarrow W\rightarrow T$ in an additive category compose to identity morphism of $T$, then $T$ is a direct summand of $W$? (2) Is it true that in an additive category if the kernel of a morphism $W\rightarrow T$ exists and $T$ is a direct summand of $W$, then $W\cong\operatorname{ker}\oplus T$? (3) Why not use a more direct proof as follows: one can show that $(**)$ is even a distinguished triangle. Also show that $(***)$ is a candidate triangle. Then the identity map between candidate triangle $(*)$ and direct sum of $(**)$ and $(***)$ is an isomorphism and we are done. Is there a problem with this proof?


(1) It's not true in general, but it is here, because of ...

(2) Yes, if by "$T$ is a direct summand of $W$", you mean that the morphism $W\to T$ is projection onto a direct summand.

More precisely, if $\varphi:W\to T$ is a morphism with a kernel, and there is another morphism $\psi:T\to W$ such that $\varphi\circ\psi=\mbox{id}_T$, then the natural map $\ker(\varphi)\to W$ and $\psi$ induce an isomorphism $\ker(\varphi)\oplus T\cong W$.

(3) The middle square won't commute if you take the identity maps (unless $\alpha=0=\beta$), so you won't get a map of triangles.

  • $\begingroup$ Thank you for your response. $\endgroup$ – Must Feb 5 '14 at 2:19

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