Lemma 12.10.7. Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. Let $\mathcal{C} \subset \mathcal{A}$ be a Serre subcategory contained in the kernel of $F$. Then $\mathcal{C} = \mathop{\mathrm{Ker}}(F)$ if and only if the induced functor $\overline{F} : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ (Lemma 12.10.6) is faithful.

Proof. We will use the results of Lemma 12.10.6 without further mention. The “only if” direction is true because the kernel of $\overline{F}$ is zero by construction. Namely, if $f : X \to Y$ is a morphism in $\mathcal{A}/\mathcal{C}$ such that $\overline{F}(f) = 0$, then $\overline{F}(\mathop{\mathrm{Im}}(f)) = \mathop{\mathrm{Im}}(\overline{F}(f)) = 0$, hence $\mathop{\mathrm{Im}}(f) = 0$ by the assumption on the kernel of $F$. Thus $f = 0$.

For the “if” direction, let $X$ be an object of $\mathcal{A}$ such that $F(X) = 0$. Then $\overline{F}(\text{id}_ X) = \text{id}_{\overline{F}(X)} = 0$, thus $\text{id}_ X = 0$ in $\mathcal{A}/\mathcal{C}$ by faithfulness of $\overline{F}$. Hence $X = 0$ in $\mathcal{A}/\mathcal{C}$, that is $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. $\square$

## Comments (2)

Comment #5396 by ykm on

minor suggestion: in statement of lemma 06XK specify what C is - i assume hypothesis from the previous lemma that C is a serre subcategory of A carries over. i was initially confused whether C=ker(F) was a definition of C.

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