## 12.9 Serre subcategories

In [Chapter I, Section 1, Serre_homotopie_classes] a notion of a “class” of abelian groups is defined. This notion has been extended to abelian categories by many authors (in slightly different ways). We will use the following variant which is virtually identical to Serre's original definition.

Definition 12.9.1. Let $\mathcal{A}$ be an abelian category.

1. A Serre subcategory of $\mathcal{A}$ is a nonempty full subcategory $\mathcal{C}$ of $\mathcal{A}$ such that given an exact sequence

$A \to B \to C$

with $A, C \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then also $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

2. A weak Serre subcategory of $\mathcal{A}$ is a nonempty full subcategory $\mathcal{C}$ of $\mathcal{A}$ such that given an exact sequence

$A_0 \to A_1 \to A_2 \to A_3 \to A_4$

with $A_0, A_1, A_3, A_4$ in $\mathcal{C}$, then also $A_2$ in $\mathcal{C}$.

In some references the second notion is called a “thick” subcategory and in other references the first notion is called a “thick” subcategory. However, it seems that the notion of a Serre subcategory is universally accepted to be the one defined above. Note that in both cases the category $\mathcal{C}$ is abelian and that the inclusion functor $\mathcal{C} \to \mathcal{A}$ is a fully faithful exact functor. Let's characterize these types of subcategories in more detail.

Lemma 12.9.2. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C}$ be a subcategory of $\mathcal{A}$. Then $\mathcal{C}$ is a Serre subcategory if and only if the following conditions are satisfied:

1. $0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$,

2. $\mathcal{C}$ is a strictly full subcategory of $\mathcal{A}$,

3. any subobject or quotient of an object of $\mathcal{C}$ is an object of $\mathcal{C}$,

4. if $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is an extension of objects of $\mathcal{C}$ then also $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

Moreover, a Serre subcategory is an abelian category and the inclusion functor is exact.

Proof. Omitted. $\square$

Lemma 12.9.3. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C}$ be a subcategory of $\mathcal{A}$. Then $\mathcal{C}$ is a weak Serre subcategory if and only if the following conditions are satisfied:

1. $0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$,

2. $\mathcal{C}$ is a strictly full subcategory of $\mathcal{A}$,

3. kernels and cokernels in $\mathcal{A}$ of morphisms between objects of $\mathcal{C}$ are in $\mathcal{C}$,

4. if $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is an extension of objects of $\mathcal{C}$ then also $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

Moreover, a weak Serre subcategory is an abelian category and the inclusion functor is exact.

Proof. Omitted. $\square$

Lemma 12.9.4. Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. Then the full subcategory of objects $C$ of $\mathcal{A}$ such that $F(C) = 0$ forms a Serre subcategory of $\mathcal{A}$.

Proof. Omitted. $\square$

Definition 12.9.5. Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. Then the full subcategory of objects $C$ of $\mathcal{A}$ such that $F(C) = 0$ is called the kernel of the functor $F$, and is sometimes denoted $\mathop{\mathrm{Ker}}(F)$.

Any Serre subcategory of an abelian category is the kernel of an exact functor. In Examples, Section 104.69 we discuss this for Serre's original example of torsion groups.

Lemma 12.9.6. 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 4.26.1.

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 4.26.20.)

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 4.26.8. $\square$

Lemma 12.9.7. Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. Then $\mathcal{C} = \mathop{\mathrm{Ker}}(F)$ if and only if the induced functor $\overline{F} : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ is faithful.

Proof. 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 $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. $\square$

Comment #2613 by Xiaofa Chen on

Why is the functor G an exact functor? Is every exact sequence in the localization category isomorphisc to some standard exact sequence? I quite doubt that.

Comment #2615 by on

@Xiaofa: Do not understand your question. The only functor $G$ on the page is a functor which is assumed to be exact!

@Everybody: If you make a comment on a lemma, proposition, definition, etc, please, please go to the tags page of the thing you want to comment on and do not comment on the webpage of the section!

Comment #3824 by Xiaofa Chen on

Yes. It's my mistake. At that time, I just learned this theory and I took it for granted that it is universal among all functors which annihilate $\mathcal C$. Then I realized that I is universal among exact functors which annihilate $\mathcal C$. And I don't know how to delete a 'stupid comment' as that.

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