The Stacks project

Proof. Additive functors commute with direct sums by Lemma 12.3.7 hence (1) implies (2) and (3). On the other hand (2) and (3) are equivalent because the composition $F(A) \oplus F(B) \to F(A \oplus B) \to F(A) \oplus F(B)$ is the identity map. Assume (2) and (3) hold. Let $f, g : A \to B$ be maps. Then $f + g$ is equal to the composition

Apply the functor $F$ and consider the following diagram

We claim this is commutative. For the middle square we can verify it separately for each of the four induced maps $F(A) \to F(B)$ where it follows from the fact that $F$ is a functor (in other words this square commutes even if $F$ does not satisfy any properties beyond being a functor). For the triangle on the left, we use that $F(A \oplus A) \to F(A) \oplus F(A)$ is an isomorphism to see that it suffice to check after composition with this map and this check is trivial. Dually for the other triangle. Thus going around the bottom is equal to $F(f + g)$ and we conclude. $\square$

Proof. If $F$ is left exact, i.e., $F$ commutes with finite limits, then $F$ sends products to products, hence $F$ preserved direct sums, hence $F$ is additive by Lemma 12.7.1. On the other hand, suppose that for every short exact sequence $0 \to A \to B \to C \to 0$ the sequence $0 \to F(A) \to F(B) \to F(C)$ is exact. Let $A, B$ be two objects. Then we have a short exact sequence

see for example Lemma 12.3.10. By assumption, the lower row in the commutative diagram

is exact. Hence by the snake lemma (Lemma 12.5.17) we conclude that $F(A) \oplus F(B) \to F(A \oplus B)$ is an isomorphism. Hence $F$ is additive in this case as well. Thus for the rest of the proof we may assume $F$ is additive.

Denote $f : B \to C$ a map from $B$ to $C$. Exactness of $0 \to A \to B \to C$ just means that $A = \mathop{\mathrm{Ker}}(f)$. Clearly the kernel of $f$ is the equalizer of the two maps $f$ and $0$ from $B$ to $C$. Hence if $F$ commutes with limits, then $F(\mathop{\mathrm{Ker}}(f)) = \mathop{\mathrm{Ker}}(F(f))$ which exactly means that $0 \to F(A) \to F(B) \to F(C)$ is exact.

Conversely, suppose that $F$ is additive and transforms any short exact sequence $0 \to A \to B \to C \to 0$ into an exact sequence $0 \to F(A) \to F(B) \to F(C)$. Because it is additive it commutes with direct sums and hence finite products in $\mathcal{A}$. To show it commutes with finite limits it therefore suffices to show that it commutes with equalizers. But equalizers in an abelian category are the same as the kernel of the difference map, hence it suffices to show that $F$ commutes with taking kernels. Let $f : A \to B$ be a morphism. Factor $f$ as $A \to I \to B$ with $f' : A \to I$ surjective and $i : I \to B$ injective. (This is possible by the definition of an abelian category.) Then it is clear that $\mathop{\mathrm{Ker}}(f) = \mathop{\mathrm{Ker}}(f')$. Also $0 \to \mathop{\mathrm{Ker}}(f') \to A \to I \to 0$ and $0 \to I \to B \to B/I \to 0$ are short exact. By the condition imposed on $F$ we see that $0 \to F(\mathop{\mathrm{Ker}}(f')) \to F(A) \to F(I)$ and $0 \to F(I) \to F(B) \to F(B/I)$ are exact. Hence it is also the case that $F(\mathop{\mathrm{Ker}}(f'))$ is the kernel of the map $F(A) \to F(B)$, and we win.

The proof of (3) is similar to the proof of (2). Statement (4) is a combination of (2) and (3). $\square$

Proof. Omitted. $\square$

Proof. As $\mathcal{B}$ is abelian we see that all finite limits and colimits exist in $\mathcal{B}$ by Lemma 12.5.5. Since $b$ is a left adjoint we see that $b$ is also right exact and hence exact, see Categories, Lemma 4.24.6. Let $\varphi : B_1 \to B_2$ be a morphism of $\mathcal{B}$. In particular, if $K = \mathop{\mathrm{Ker}}(B_1 \to B_2)$, then $K$ is the equalizer of $0$ and $\varphi $ and hence $bK$ is the equalizer of $0$ and $b\varphi $, hence $bK$ is the kernel of $b\varphi $. Similarly, if $Q = \mathop{\mathrm{Coker}}(B_1 \to B_2)$, then $Q$ is the coequalizer of $0$ and $\varphi $ and hence $bQ$ is the coequalizer of $0$ and $b\varphi $, hence $bQ$ is the cokernel of $b\varphi $. Thus we see that every morphism of the form $b\varphi $ in $\mathcal{A}$ has a kernel and a cokernel. However, since $ba \cong \text{id}$ we see that every morphism of $\mathcal{A}$ is of this form, and we conclude that kernels and cokernels exist in $\mathcal{A}$. In fact, the argument shows that if $\psi : A_1 \to A_2$ is a morphism then

Now we still have to show that $\mathop{\mathrm{Coim}}(\psi )= \mathop{\mathrm{Im}}(\psi )$. We do this as follows. First note that since $\mathcal{A}$ has kernels and cokernels it has all finite limits and colimits (see proof of Lemma 12.5.5). Hence we see by Categories, Lemma 4.24.6 that $a$ is left exact and hence transforms kernels (=equalizers) into kernels.

Thus the lemma holds. $\square$