The motivation is to show that $N_1,N_2$ are flat $R$-module if and only if $N_1\oplus N_2$ is flat $R$-module. We use the following fact:

Given two sequences of $R$-modules: $$0\rightarrow M_1'\rightarrow M_1\rightarrow M_1''\rightarrow 0$$ and $$0\rightarrow M_2'\rightarrow M_2\rightarrow M_2''\rightarrow 0$$ are exact if and only if $$0\rightarrow M_1'\oplus M_2'\rightarrow M_1\oplus M_2\rightarrow M_1''\oplus M_2''\rightarrow 0$$ is exact.

The proof is just the routine check.

Now I was wondering if this statement is still true for any category, not just for the category of $R$-modules. If so, is there a general (categorical) proof for the above statement?

More generally, if we replace the $I=\{1,2\}$ by an arbitrary index set, is this still true? (We know for the category of $R$-modules, it is true.)


A categorical proof could go as follows (for $I=\{1,2\}$ and then of course induction yields the finite $I$ cases. As Kevin Carlson points out, this will fail in general for arbitrary $I$)

Short version: finite direct sums in an abelian (or additive, provided the direct sums in question exist -one may even require less) category are both a coproduct and a product, so both a colimit and a limit, hence they commute both with cokernels (because colimits commute with colimits) and with kernels (because limits commute with limits). Hence they preserve exact sequences.

Long(er) version: The fact that the new sequence is a complex is clear (just use bilinearity of composition in an additive/abelian category, or in a more general setting of, say punctured categories, just use the fact that $(f_1\oplus f_2)\circ (g_1\oplus g_2) = (f_1\circ g_1) \oplus (f_2\circ g_2)$ by some abstract nonsense about coproducts, and then that $0\oplus 0 = 0$ by some abstract nonsense in punctured categories), so we only need to show that it is exact.

Let $x\in_U M_1\oplus M_2$ be a generalized element (that is a map $x: U\to M_1\oplus M_2$) such that $x$ is mapped to $0$ by $M_1\oplus M_2 \to M_1''\oplus M_2''$. Since $M_1\oplus M_2$ is the product of $M_1,M_2$, we may write $x$ as $(y,z)$.But then $y$ is sent to $0$ by $M_1\to M_1''$ and similarly for $z$. Hence $y$ factorizes uniquely through $M_1'$ and $z$ uniquely through $M_2'$, by exactness of the previous sequences. Hence $x$ factors uniquely through $M_1'\oplus M_2'$. Thus $M_1'\oplus M_2'\to M_1\oplus M_2$ is the kernel of $M_1\oplus M_2\to M_1''\oplus M_2''$ (in particular it is a monomorphism).

The only thing left to check is that the sequence is exact at $M_1''\oplus M_2''$, that is $M_1\oplus M_2\to M_1''\oplus M_2''$ needs to be epi. But $\oplus$ is a coproduct, and coproducts preserve epis, so we're good.

Notice that we have used that $\oplus$ was a coproduct and a product : this is why everything fails for an infinite $I$: an infinite direct sum is not a product.


The answers to your questions are, respectively, no, yes, and no.

  1. The statement is true for arbitrary abelian categories, but not all categories, since not all categories even have things like exact sequences.
  2. A categorical proof of this presumably exists, but as with the snake lemma, you might as well use the Freyd-Mitchell embedding theorem and reduce it to the case of modules.
  3. Replacing $I$ with arbitrary index sets won't work because not all abelian categories have infinite direct sums - for example, the category of finite abelian groups. That said, if all the direct sums in question do exist, the proof should go through pretty much the same way. EDIT: It does not, see comments.
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    $\begingroup$ 3. is not correct. Infinite direct sums need not preserve monomorphisms. This failure is very common in the opposites of categories of sheaves of abelian groups. In fact, axiom AB4 (so beyond the axioms of abelian categories) in Grothendieck's Tohoku paper is precisely that infinite direct sums (exist and) are exact. $\endgroup$ – Kevin Carlson Jul 4 '18 at 5:29
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    $\begingroup$ Thank you for that Carlson $\endgroup$ – Cory Griffith Jul 4 '18 at 17:44

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