The Stacks project

Proof. Consider the set of arrows of $\mathcal{A}$ defined by the following formula

We claim that $S$ is a multiplicative system. To prove this we have to check MS1, MS2, MS3, see Categories, Definition 4.27.1.

It is clear that identities are elements of $S$. Suppose that $f : A \to B$ and $g : B \to C$ are elements of $S$. There are exact sequences

Hence it follows that $gf \in S$. This proves MS1. (In fact, a similar argument will show that $S$ is a saturated multiplicative system, see Categories, Definition 4.27.20.)

Consider a solid diagram

with $t \in S$. Set $W = C \amalg _ A B = \mathop{\mathrm{Coker}}((t, -g) : A \to C \oplus B)$. Then $\mathop{\mathrm{Ker}}(t) \to \mathop{\mathrm{Ker}}(s)$ is surjective and $\mathop{\mathrm{Coker}}(t) \to \mathop{\mathrm{Coker}}(s)$ is an isomorphism. Hence $s$ is an element of $S$. This proves LMS2 and the proof of RMS2 is dual.

Finally, consider morphisms $f, g : B \to C$ and a morphism $s : A \to B$ in $S$ such that $f \circ s = g \circ s$. This means that $(f - g) \circ s = 0$. In turn this means that $I = \mathop{\mathrm{Im}}(f - g) \subset C$ is a quotient of $\mathop{\mathrm{Coker}}(s)$ hence an object of $\mathcal{C}$. Thus $t : C \to C' = C/I$ is an element of $S$ such that $t \circ (f - g) = 0$, i.e., such that $t \circ f = t \circ g$. This proves LMS3 and the proof of RMS3 is dual.

Having proved that $S$ is a multiplicative system we set $\mathcal{A}/\mathcal{C} = S^{-1}\mathcal{A}$, and we set $F$ equal to the localization functor $Q$. By Lemma 12.8.4 the category $\mathcal{A}/\mathcal{C}$ is abelian and $F$ is exact. If $X$ is in the kernel of $F = Q$, then by Lemma 12.8.3 we see that $0 : X \to Z$ is an element of $S$ and hence $X$ is an object of $\mathcal{C}$, i.e., the kernel of $F$ is $\mathcal{C}$. Finally, if $G$ is as in the statement of the lemma, then $G$ turns every element of $S$ into an isomorphism. Hence we obtain the functor $H : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ from the universal property of localization, see Categories, Lemma 4.27.8. We still have to show the functor $H$ is exact. To do this it suffices to show that $H$ commutes with taking kernels and cokernels, see Lemma 12.7.2. Let $A \to B$ be a morphism in $\mathcal{A}/\mathcal{C}$. We may represent $A \to B$ as $fs^{-1}$ where $s : A' \to A$ is in $S$ and $f : A' \to B$ an arbitrary morphism of $\mathcal{A}$. Since $F = Q$ maps $s$ to an isomorphism in the quotient category $\mathcal{A}/\mathcal{C}$, it suffices to show that $H$ commutes with taking kernels and cokernels of morphisms $f : A \to B$ of $\mathcal{A}$. But here we have $H(f) = G(f)$ and the result follows from the fact that $G$ is exact. $\square$