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?
 A: 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.
A: 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 $H = \operatorname{im}(\eta)$.

*

*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.)


*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:

*

*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.


*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.


*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.


*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 $\mathscr G$.
