nLab
additive functor

A functor $F:C\to D$ between additive categories is itself additive if it preserves finite biproducts.

That is, $F$ maps a zero object to a zero object; and, given objects $x$ and $y$ of $C$, not only is $F(x\oplus y)\cong F(x)\oplus F(y)$, but $F$ even preserves the inclusion and projection maps:

$$\begin{array}{ccccc}x& & & & y\\ & \searrow & & \swarrow \\ & & x\oplus y\\ & \swarrow & & \searrow \\ x& & & & y\end{array}\stackrel{F}{\mapsto }\begin{array}{ccccc}F(x)& & & & y\\ & \searrow & & \swarrow \\ & & F(x\oplus y)\cong F(x)\oplus F(y)\\ & \swarrow & & \searrow \\ F(x)& & & & y\end{array}$$\array { x & & & & y \\ & \searrow & & \swarrow \\ & & x \oplus y \\ & \swarrow & & \searrow \\ x & & & & y } \quad\quad\stackrel{F}{\mapsto}\quad\quad \array { F(x) & & & & y \\ & \searrow & & \swarrow \\ & & F(x \oplus y) \cong F(x) \oplus F(y) \\ & \swarrow & & \searrow \\ F(x) & & & & y }

Additive categories are always enriched over Ab, and an additive functor is always an enriched functor accordingly. This need not be stated as a requirement; it follows from preserving biproducts, since the $\mathrm{Ab}$-enrichment structure may be recovered from the biproducts (as described at biproduct). Conversely, any $\mathrm{Ab}$-enriched functor automatically preserves any finite biproducts that may exist, since finite biproducts in Ab-enriched categories are Cauchy colimits.

In practice, functors between additive categories are generally assumed to be additive.