The Stacks project

Lemma 12.3.15. Let $\mathcal{A}$ be a preadditive category, $x$ be an object of $\mathcal{A}$, and $f : x \to x$ be idempotent. Write $g = \text{id} _ x - f$ and recall by Remark 12.3.14 that $f \circ g = g \circ f = 0$.

  1. If $f$ has kernel $i : \mathop{\mathrm{Ker}}(f) \to x$ and $p$ is the unique morphism such that $g = i \circ p$, then $p \circ i = \text{id} _{\mathop{\mathrm{Ker}}(f)}$ and $p$ is the cokernel of $f$.

  2. If $f$ has cokernel $p : x \to \mathop{\mathrm{Coker}}(f)$ and $i$ is the unique morphism such that $g = i \circ p$, then $p \circ i = \text{id} _{\mathop{\mathrm{Coker}}(f)}$ and $i$ is the kernel of $f$.

  3. If $g$ has kernel $j : \mathop{\mathrm{Ker}}(g) \to x$ and $q$ is the unique morphism such that $f = j \circ q$, then $q \circ j = \text{id} _{\mathop{\mathrm{Ker}}(g)}$ and $q$ is the cokernel of $g$.

  4. If $g$ has cokernel $q : x \to \mathop{\mathrm{Coker}}(g)$ and $j$ is the unique morphism such that $f = j \circ q$, then $q \circ j = \text{id} _{\mathop{\mathrm{Coker}}(g)}$ and $j$ is the kernel of $g$.

Proof. As for (1), compute first that $i = (f + g) \circ i = g \circ i$. By the construction of $p$, this implies that $i \circ p \circ i = g \circ i = i \circ \text{id} _{\mathop{\mathrm{Ker}}(f)}$. By Lemma 12.3.11 we conclude that $p \circ i = \text{id} _{\mathop{\mathrm{Ker}}(f)}$.

It remains to show that $p : x \to \mathop{\mathrm{Ker}}(f)$ satisfies the universal property of the cokernel of $f$, i.e., that given a morphism $a : x \to y$ satisfying $a \circ f = 0$ there exists a unique morphism $b : \mathop{\mathrm{Ker}}(f) \to y$ such that $a = b \circ p$. Existence follows by first computing that $a = a \circ (f + g) = a \circ g$ and concluding by the construction of $p$ that $a = a \circ g = a \circ i \circ p$ (so that we can take $b = a \circ i$). Uniqueness follows from that $b = b \circ p \circ i = a \circ i$ (so that $a$ determines $b$).

By duality, symmetry (i.e., that $g$ is itself idempotent; cf. Remark 12.3.14), and duality and symmetry respectively, (2), (3), and (4) follow as did (1). $\square$

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