Lemma 12.3.17. Let $\mathcal{A}$ be a preadditive category, $x$ and $y$ be objects of $\mathcal{A}$, and $j : y \to x$ and $q : x \to y$ be morphisms satisfying $q \circ j = \text{id} _ y$. Then
$\text{id} _ x - j \circ q$ has kernel $j$.
$\text{id} _ x - j \circ q$ has cokernel $q$.
Proof. As for (1), compute first that $(\text{id} _ x - j \circ q) \circ j = 0$. We must show that $j$ satisfies the universal property of the kernel of $\text{id} _{x} - j \circ q$, i.e., that given a morphism $a : z \to x$ satisfying $(\text{id} _ x - j \circ q) \circ a = 0$ there exists a unique morphism $b : z \to y$ such that $a = j \circ b$. Existence follows by computing that $a = j \circ q \circ a$ (so that we can take $b = q \circ a$). Uniqueness follows from that $b = q \circ j \circ b = q \circ a$ (so that $a$ determines $b$).
By duality, (2) follows as did (1). $\square$
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