Additive, covariant functor preserve direct sum?

Question

Does any additive, covariant functor preserve direct sum?

Answer 1

I'm not sure what is being asked, but I'll try rephrasing what I suspect the question is: if $A$ and $B$ are enriched in abelian groups, and if $F: A \to B$ is a functor such that all the maps $\hom(a, a') \to \hom(F(a), F(a'))$ are homomorphisms, then is it true that $F$ takes finite sums (coproducts) in $A$ to finite sums in $B$? The answer to that question is 'yes'. José Figueroa-O'Farrill has already given some hints, but I think they deserve to be amplified.

The result is standard in homological algebra textbooks (I believe you can find it in Hilton and Stammbach for instance). It helps to observe first that in $Ab$-enriched categories $A$, or even in categories enriched in commutative monoids, finite sums in $A$ are the same as biproducts, which roughly speaking are simultaneously products and coproducts (in a compatible way).

Suppose $a + b$ is a coproduct in $A$. Then by definition there are coproduct inclusions $i_a: a \to a + b$, $i_b: b \to a + b$. Also by definition, a map $f: a + b \to c$ is uniquely determined by the pair of maps $f_a := f \circ i_a$, $f_b = f \circ i_b$. In particular, we have a map $p_a: a + b \to a$ determined by the pair $1_a: a \to a$, $0: b \to a$, where $0$ denotes the zero element in the abelian group $\hom(b, a)$. Similarly, we have a map $p_b: a + b \to b$. By definition, we have the equations

$$p_a i_a = id_a \qquad p_a i_b = 0 \qquad p_b i_a = 0 \qquad p_b i_b = id_b$$

Note also that $i_a p_a + i_b p_b = 1_{a+b}$, because each side is a map $f$ satisfying the equations $f i_a = i_a$, $f i_b = i_b$, and there is only one such map (by the universal property of coproducts).

Now, the crucial point is that whenever you have maps $i_a: a \to c$, $i_b: b \to c$, $p_a: c \to a$, $p_b: c \to b$ satisfying the four equations above and a fifth equation $i_a p_a + i_b p_b = 1_c$, then $c$ is simultaneously a product and coproduct of $a$ and $b$; in this situation we call $c$ a biproduct. I invite you to write out the proof. (Hint: given $g: d \to a$ and $h: d \to b$, define $\langle g, h \rangle: d \to c$ to be $i_a g + i_b h$. Show this is the unique map such that $p_a \langle g, h \rangle = g$ and $p_b \langle g, h \rangle = h$, so that $c$ satisfies the universal property of a product. By dualizing this argument, $c$ also satisfies the universal property of a coproduct.)

But any $Ab$-enriched functor preserves these five equations since it preserves composition, identities, zero elements, and addition. Therefore $Ab$-enriched functors $F: A \to B$ take biproducts in $A$ to biproducts in $B$. That is the same as saying it takes coproducts in $A$ to coproducts in $B$.

Just to round out the story: there are converses to these statements as follows. If a category $A$ has biproducts (see the nLab page cited above), then $A$ is automatically enriched in commutative monoids: the sum of two morphisms $f, g \in hom(a, b)$ is given by the composite

$$a \stackrel{\Delta_a}{\to} a \oplus a \stackrel{f \oplus g}{\to} b \oplus b \stackrel{\nabla_b}{\to} b$$

where $\oplus$ denotes biproducts, and the diagonal and codiagonal are defined using the the biproduct structure. (The zero element in $\hom(a, b)$ is the composite $a \to 0 \to b$ factoring through the zero object.) And if $A$ and $B$ have biproducts and $F: A \to B$ preserves them, then $F$ is an enriched functor: preserves the addition and zero carried by the hom-sets.