Lemma 12.7.2. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.

1. If $F$ is either left or right exact, then it is additive.

2. $F$ is left exact if and only if 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.

3. $F$ is right exact if and only if for every short exact sequence $0 \to A \to B \to C \to 0$ the sequence $F(A) \to F(B) \to F(C) \to 0$ is exact.

4. $F$ is exact if and only if for every short exact sequence $0 \to A \to B \to C \to 0$ the sequence $0 \to F(A) \to F(B) \to F(C) \to 0$ is exact.

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

$0 \to A \to A \oplus B \to B \to 0$

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

$\xymatrix{ 0 \ar[r] & F(A) \ar[d] \ar[r] & F(A) \oplus F(B) \ar[r] \ar[d] & F(B) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & F(A) \ar[r] & F(A \oplus B) \ar[r] & F(B) }$

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). $\square$

## Comments (0)

There are also:

• 3 comment(s) on Section 12.7: Additive functors

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 010N. Beware of the difference between the letter 'O' and the digit '0'.