# Is every additive monofunctor between abelian categories left exact?

Is there an additive functor between abelian categories, which preserves monomorphisms, but is not left exact?

Consider an integral domain $R$ which is not a field, and choose $a \in R \setminus (\{0\} \cup R^*)$. Consider the functor $\mathsf{Mod}(R) \to \mathsf{Mod}(R)$ which maps an $R$-module $M$ to its submodule $aM$. Clearly this preserves monomorphisms, but it is not left exact: The exact sequence $0 \to R \to Q(R) \to Q(R)/R$ becomes $0 \to aR \to Q(R) \to Q(R)/R$ after applying the functor.

Alternatively, it suffices to give an additive functor between abelian categories which preserves epimorphisms, but is not right exact. Take an integral domain $R$, which is not a field, and consider the functor $\mathsf{Mod}(R) \to \mathsf{Mod}(R)$ which takes an $R$-module to its maximal torsionfree quotient, i.e. which mods out the torsion submodule. This functor preserves epimorphisms, but is not right exact: The exact sequence $R \to Q(R) \to Q(R)/R \to 0$ becomes $R \to Q(R) \to 0 \to 0$ after applying the functor.

• Actually, both your examples work on both sides: they preserve both monomorphisms and epimorphisms, but are neither left exact nor right exact. Dec 3, 2019 at 4:57

Here is a construction that generalises both examples in Martin Brandenburg's answer:

Lemma. Let $$F, G \colon \mathscr A \to \mathscr B$$ be additive functors between abelian categories, let $$\eta \colon F \to G$$ be a natural transformation, and let $$H = \operatorname{im}(\eta)$$.

1. The functor $$H$$ is additive. It preserves monomorphisms if $$G$$ does, and it preserves epimorphisms if $$F$$ does. (For example, these hypotheses are satisfied when $$F$$ is right exact and $$G$$ is left exact.)

2. If $$F$$ and $$G$$ are exact, then the following are equivalent:

i. $$H$$ is exact in the middle;

ii. $$H$$ is exact;

iii. $$\ker(\eta)$$ is right exact;

iv. $$\operatorname{coker}(\eta)$$ is left exact.

Proof. For any $$A, B \in \mathscr A$$, we get a commutative diagram $$\begin{array}{ccc} F(A) \oplus F(B) & \twoheadrightarrow & H(A) \oplus H(B) & \hookrightarrow & G(A) \oplus G(B) \\ \wr\downarrow\ & & \downarrow & & \wr\downarrow\ \\ F(A \oplus B) & \twoheadrightarrow & H(A \oplus B) & \hookrightarrow & G(A \oplus B).\! \end{array}$$ We conclude that the middle vertical arrow is both monic and epic, hence an isomorphism. Now let $$A \to B$$ be a monomorphism in $$\mathscr A$$. If $$G$$ preserves monomorphisms, then the bottom and vertical arrows in the commutative diagram $$\begin{array}{ccc}H(A) & \to & H(B) \\ \downarrow & & \downarrow \\ G(A) & \to & G(B) \end{array}$$ are monomorphisms, hence so is the top arrow. The statement about epimorphisms follows dually, finishing the proof of (1).

In (2), the equivalence i $$\Leftrightarrow$$ ii is clear from (1), as a middle exact functor that preserves monomorphisms and epimorphisms is exact. Write $$K = \ker(\eta)$$ and $$Q = \operatorname{coker}(\eta)$$. Let $$0 \to A \to B \to C \to 0$$ be a short exact sequence, giving a commutative diagram with exact rows $$\begin{array}{ccccccccc} 0 & \to & F(A) & \to & F(B) & \to & F(C) & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & G(A) & \to & G(B) & \to & G(C) & \to & 0.\! \end{array}$$ By the snake lemma, we see that $$K$$ is left exact and $$Q$$ is right exact. If $$K$$ is moreover right exact, applying the same reasoning to $$K \hookrightarrow F$$ shows that $$H = \operatorname{coker}(K \to F)$$ is right exact, hence exact since it preserves monomorphisms (this also follows directly from the snake lemma since $$K \to F$$ is injective). This proves iii $$\Rightarrow$$ ii, and iv $$\Rightarrow$$ ii follows dually. Conversely, if $$H$$ is exact, applying the same reasoning to $$F \twoheadrightarrow H$$ shows that $$K$$ is right exact, and dually that $$Q$$ is left exact. $$\square$$

Examples. There are many interesting examples where $$F$$ and $$G$$ are right and left exact respectively:

1. Let $$R$$ be a domain that is not a field, and let $$a \in R$$ neither zero nor a unit. Let $$F = G = \operatorname{id} \colon \mathbf{Mod}_R \to \mathbf{Mod}_R,$$ and $$\eta \colon F \to G$$ multiplication by $$a$$. Then $$\operatorname{im}(\eta)(M) = aM$$, which is an additive functor preserving monomorphisms and epimorphisms, but it is not exact since $$M \mapsto M/aM = M \otimes_R R/(a)$$ is not left exact.

2. Let $$R$$ be a domain that is not a field and let $$Q$$ be its field of fractions. Let $$F = \operatorname{id} \cong (-) \otimes_R R$$ and $$G = (-) \otimes_R Q$$, and $$\eta \colon F \to G$$ the natural transformation induced by $$R \to Q$$. Then $$F$$ and $$G$$ are exact, so $$\operatorname{im}(\eta)(M) \cong M/M_{\operatorname{tors}}$$ is additive and preserves monomorphisms and epimorphisms. But it is not exact, since $$\ker(\eta) = (-)_{\operatorname{tors}}$$ is not right exact, or since $$\operatorname{coker}(\eta) = (-) \otimes_R Q/R$$ is not left exact.

3. Let $$P = [1]$$ be the poset $$\{0 < 1\}$$, and let $$\mathscr A = [P,\mathbf{Ab}]$$. Then $$\mathscr A$$ comes with exact evaluation functors $$\operatorname{ev}_i \colon \mathscr A \to \mathbf{Ab}$$ as well as a natural transformation $$\eta \colon \operatorname{ev}_0 \to \operatorname{ev}_1$$ induced by the morphism $$0 < 1$$ in $$P$$. This is the example given here of an additive functor that is not exact in the middle.

4. Let $$X$$ be a topological space on which there exists a separated presheaf $$\mathscr F$$ that is not a sheaf. This is the case if and only if the opens in $$X$$ are not linearly ordered: as soon as $$U = U_1 \cup U_2$$ with $$U_1, U_2 \subsetneq U$$, then the presheaf image $$\mathscr F$$ of $$\mathbf Z_{U_1} \oplus \mathbf Z_{U_2} \to \mathbf Z_U$$ is separated (being a subpresheaf of a sheaf), but the local sections $$1$$ on $$U_1$$ and $$U_2$$ are in $$\mathscr F$$ whereas their glueing to $$U$$ is not.

Let $$\mathscr A = \mathscr B = \mathbf{PSh}(X)$$ and $$F = \operatorname{id}$$ and $$G = \iota(-)^\#$$, and $$\eta \colon F \to G$$ the unit of the adjunction $$(-)^\# \colon \mathbf{PSh}(X) \leftrightarrows \mathbf{Sh}(X)\,\colon\!\iota$$. Then $$F$$ is exact and $$G$$ is left exact, so by the lemma the image $$H = \operatorname{im}(\eta)$$ is additive and preserves monomorphisms and epimorphisms.

For a separated presheaf $$\mathscr F$$ that is not a sheaf, consider the short exact sequence $$0 \to \mathscr F \to \iota\mathscr F^\# \to \mathscr G \to 0$$ of presheaves. Then $$\mathscr G^\# = 0$$ by exactness of sheafification, so applying $$H$$ gives $$0 \to \mathscr F \to \iota\mathscr F^\# \to 0 \to 0,$$ which is not exact. Alternatively, $$K = \ker(\eta)$$ is zero on $$\mathscr F$$ and $$\iota\mathscr F^\#$$, but not on $$\mathscr G$$.