Lemma 13.4.12. Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent

1. $f$ has a kernel,

2. $f$ has a cokernel,

3. $f$ is the isomorphic to a composition $K \oplus Z \to Z \to Z \oplus Q$ of a projection and coprojection for some objects $K, Z, Q$ of $\mathcal{D}$.

Proof. Any morphism isomorphic to a map of the form $X' \oplus Z \to Z \oplus Y'$ has both a kernel and a cokernel. Hence (3) $\Rightarrow$ (1), (2). Next we prove (1) $\Rightarrow$ (3). Suppose first that $f : X \to Y$ is a monomorphism, i.e., its kernel is zero. By TR1 there exists a distinguished triangle $(X, Y, Z, f, g, h)$. By Lemma 13.4.1 the composition $f \circ h[-1] = 0$. As $f$ is a monomorphism we see that $h[-1] = 0$ and hence $h = 0$. Then Lemma 13.4.11 implies that $Y = X \oplus Z$, i.e., we see that (3) holds. Next, assume $f$ has a kernel $K$. As $K \to X$ is a monomorphism we conclude $X = K \oplus X'$ and $f|_{X'} : X' \to Y$ is a monomorphism. Hence $Y = X' \oplus Y'$ and we win. The implication (2) $\Rightarrow$ (3) is dual to this. $\square$

## Comments (2)

Comment #2005 by luke on

In Lemma 13.4.11, (1) implies (3), it should be "$f \circ h[-1]$" instead of "$h[-1] \circ f = 0$".

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• 13 comment(s) on Section 13.4: Elementary results on triangulated categories

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