# The Stacks Project

## Tag 0DLP

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 for 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. $\square$

The code snippet corresponding to this tag is a part of the file homology.tex and is located in lines 1298–1310 (see updates for more information).

\begin{lemma}
Let $\mathcal{A}$ and $\mathcal{B}$ be additive categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.
The following are equivalent
\begin{enumerate}
\item $F$ is additive,
\item $F(A) \oplus F(B) \to F(A \oplus B)$ is an isomorphism for
all $A, B \in \mathcal{A}$, and
\item $F(A \oplus B) \to F(A) \oplus F(B)$  is an isomorphism for
all $A, B \in \mathcal{A}$.
\end{enumerate}
\end{lemma}

\begin{proof}
Additive functors commute with direct sums by
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 for 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.
\end{proof}

Comment #2514 by Jo on April 21, 2017 a 11:08 am UTC

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 April 21, 2017 a 12:34 pm UTC

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 #2557 by Johan (site) on May 25, 2017 a 6:01 pm UTC

Thanks to Jo and Andrea. Fixed here.

## Add a comment on tag 0DLP

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 lower-right corner).