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In my Rings and Modules class, we defined additive functors this way:

A (covariant/contravariant) functor $F$ is called additive if:

(i). $F0 = 0$, where $0$ is the zero-module.

(ii). For any modules $M, N$ we have $F(M \oplus N) = FM \oplus FN$.

I looked around online and I saw this definitions, which looks the same but uses morphisms instead of modules:

A (covariant/contravariant) functor $F$ is called additive if:

(i). $F0 = 0$, where $0$ is the zero-map.

(ii). For any modules $M, N$ and morphisms $\phi, \psi: M \longrightarrow N$ we have $F(\phi + \psi) = F \phi + F \psi : FM \longrightarrow FN$.

Now it seems clear, intuitively, how these two concepts are equivalent. However, I always have a lot of trouble proving elementary stuff with functors, because I always think if what I'm doing is allowed.

I was trying to prove that the first two conditions imply the other two. So I tried to rephrase them in terms of compositions of homomorphisms.

(i). Let $\phi : M \longrightarrow N$ be the zero-map. Then $\phi$ is the same as $\iota \circ \phi_0$, where $\phi_0 : M \longrightarrow 0$ and $\iota : 0 \longrightarrow N$. Then $F \phi = F(\iota \circ \phi_0) = F \iota \circ F \phi_0$ is the zero-map, because we are assuming that $F0 = 0$.

(ii). Let $\phi, \psi: M \longrightarrow N$ be homomorphisms. Then $\phi + \psi$ is the same as the following sequence: $$ M \overset{id_M \oplus id_M}{\longrightarrow} M \oplus M \overset{(\phi, \psi)}{\longrightarrow} N \oplus N \overset{id_N + id_N}{\longrightarrow}N.$$ At this point, however I am stuck. I don't know how to justify that applying $F$ to this sequence gives $F\phi + F\psi$, since natural stuff such as inclusions and projections are not necessarily preserved. How should I go on?

I haven't yet tried proving the other direction...

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  • $\begingroup$ "since natural stuff such as inclusions and projections are not necessarily preserved" I think they're supposed to be, and that this is implicit in what is meant by $F(M \oplus N) = FM\oplus FN$ $\endgroup$ Jan 7, 2018 at 14:38
  • $\begingroup$ math.stackexchange.com/q/2594527/404944 $\endgroup$
    – frafour
    Jan 7, 2018 at 14:50
  • $\begingroup$ math.stackexchange.com/q/2593095/404944 $\endgroup$
    – frafour
    Jan 7, 2018 at 14:50
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    $\begingroup$ Your definition of $F$ preserves sums is not complete. There are maps $F(i):FM \to F(M\oplus N)$ and $F(j):FN\to F(M\oplus N)$, and you want that $F(i)\oplus F(j): FM\oplus FN\to F(M\oplus N)$ be an isomorphism. $\endgroup$
    – Pedro
    Jan 9, 2018 at 15:35

2 Answers 2

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By definition of the sum $M\oplus N$, there exist $i_M:M\rightarrow M\oplus N, i_N:N\rightarrow M\oplus N$, such that for every morphism $f:M\rightarrow P, g:N\rightarrow P$, there exists a unique morphism $f\oplus g:M\oplus N\rightarrow P$ such that $(f\oplus g)\circ i_M=f$ and $(f\oplus g)\circ i_N=g$, take $M=N$, $f=g=Id_N$, you obtain $Id_N\oplus Id_N$. The fact that $F$ preserves sum is equivalent to saying that $F(i_N)=i_{F(N)}$ and $F(f+g)=F(f)\oplus F(g)$, this implies that $F(Id_N\oplus Id_N)=F(Id_N)\oplus F(Id_N)$, you can apply $F$ to the sequence.

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  • $\begingroup$ Thank you for your answer! I can see how this solves the problem for the first two arrows. How about the third? There we have $+$, not $\oplus$... $\endgroup$
    – frafour
    Jan 7, 2018 at 15:04
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Your first definition is imprecise. What you should say is that the universal property of coproducts furnishes a natural map $F(M) \oplus F(N) \to F(M \oplus N)$ and you want these natural maps to be isomorphisms. Anyway, see this blog post for a thorough discussion. The upshot is that in an additive category direct sums determine the abelian group structure on morphisms.

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