The Stacks project

Lemma 12.10.6. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C} \subset \mathcal{A}$ be a Serre subcategory. There exists an abelian category $\mathcal{A}/\mathcal{C}$ and an exact functor

\[ F : \mathcal{A} \longrightarrow \mathcal{A}/\mathcal{C} \]

which is essentially surjective and whose kernel is $\mathcal{C}$. The category $\mathcal{A}/\mathcal{C}$ and the functor $F$ are characterized by the following universal property: For any exact functor $G : \mathcal{A} \to \mathcal{B}$ such that $\mathcal{C} \subset \mathop{\mathrm{Ker}}(G)$ there exists a factorization $G = H \circ F$ for a unique exact functor $H : \mathcal{A}/\mathcal{C} \to \mathcal{B}$.

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

\[ S = \{ f \in \text{Arrows}(\mathcal{A}) \mid \mathop{\mathrm{Ker}}(f), \mathop{\mathrm{Coker}}(f) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) \} . \]

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

\[ \begin{matrix} 0 \to \mathop{\mathrm{Ker}}(f) \to \mathop{\mathrm{Ker}}(gf) \to \mathop{\mathrm{Ker}}(g) \\ \mathop{\mathrm{Coker}}(f) \to \mathop{\mathrm{Coker}}(gf) \to \mathop{\mathrm{Coker}}(g) \to 0 \end{matrix} \]

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

\[ \xymatrix{ A \ar[d]_ t \ar[r]_ g & B \ar@{..>}[d]^ s \\ C \ar@{..>}[r]^ f & C \amalg _ A B } \]

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. $\square$

Comments (5)

Comment #469 by Nuno on

(i) The sequence is not right exact and the sequence is not left exact. This will not affect the proof though. Also, I believe it would be instructive to point out that that S is a saturated multiplicative system. (ii) Just after the diagram, should be the cokernel of . (iii) Following this approach, is there an easy proof of the fact that if is a "big" well-powered category, then exists for any Serre subcategory ?

Comment #484 by on

Thanks for the fixes. You can find the corresponding edits here.

Unfortunately, I do not know the answer to your question at the moment. I have never worked with well powered categories.

Comment #525 by Nuno on

After some search, I've found a proof of (iii) in Popescu's Theory of Categories (theorem 7.8, chapter 4, page 312), but didn't check it yet.

Comment #7092 by Gabriel Ribeiro on

Dear Johan,

You never proved that the induced functor is exact. If this is trivial, that's fine. However, the only way that I can think about doing this is by proving that every short exact sequence in the quotient is isomorphic to the image of some short exact sequence under the quotient functor. This is not too bad but I think it would be better to at least say something.

Or do you have an easier / more elegant proof in mind?

Thanks, Gabriel

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