Lemma 12.7.1. Let $\mathcal{A}$ and $\mathcal{B}$ be additive categories. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor. The following are equivalent

1. $F$ is additive,

2. $F(A) \oplus F(B) \to F(A \oplus B)$ is an isomorphism for all $A, B \in \mathcal{A}$, and

3. $F(A \oplus B) \to F(A) \oplus F(B)$ is an isomorphism for all $A, B \in \mathcal{A}$.

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

$A \to A \oplus A \xrightarrow {\text{diag}(f, g)} B \oplus B \to B$

Apply the functor $F$ and consider the following diagram

$\xymatrix{ F(A) \ar[r] \ar[rd] & F(A \oplus A) \ar[rr]_{F(\text{diag}(f, g))} & & F(B \oplus B) \ar[r] \ar[d] & F(B) \\ & F(A) \oplus F(A) \ar[u] \ar[rr]^{\text{diag}(F(f), F(g))} & & F(B) \oplus F(B) \ar[ru] }$

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. $\square$

Comment #2514 by Jo on

It is not clear what $\text{diag}(f,g)$ is, at least to me, and I couldn't find any reference to that particular morphism anywhere on this page.

Comment #2515 by Andrea on

I think that $\text{diag}(f,g)$ is the morphism which sends an element $(a,a')$ of $A\oplus A$ to the element $(f(a),g(a'))$ of $B\oplus B$. I think also that the top central horizontal arrow of the diagram should be $F(\text{diag}(f,g))$ instead of $\text{diag}(f,g)$.

Comment #5937 by Gabriel Ribeiro on

Actually we don't need to compose with the isomorphisms to check whether the triangles commute or not. They commute by the very definition of the morphisms in (2) and (3). Also, I would suggest that it is perhaps clearer to use the remark 0103 instead of the lemma 0105 to prove that (1) implies (2) and (3).

Comment #6126 by on

@#5937. No, I don't agree. If $F$ is for example the functor $V \mapsto V \otimes V$ on vector spaces over a field, then these triangles do not commute and hence you need to use something when you check they do commute. (The funny thing about the proof is that the commutativity of the square in the diagram doesn't require $F$ to have any property.)

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