Let
$$0\longrightarrow L^{(i)}\longrightarrow M^{(i)}\longrightarrow N^{(i)}\longrightarrow 0$$
be a short exact sequence of abelian groups for every index $i$. Clearly if I take finite direct products, then
$$0\longrightarrow \prod_iL^{(i)}\longrightarrow\prod_i M^{(i)}\longrightarrow \prod_iN^{(i)}\longrightarrow 0$$
is a short exact sequence. But what about infinite direct product? Is the exactness preserved?
[Math] short exact sequences and direct product
abstract-algebrahomological-algebra
Related Solutions
Say $$0\to A'\stackrel{f}{\longrightarrow} A \stackrel{g}{\longrightarrow} A''\to 0$$ is split (and exact); then there exist $p\colon A\to A'$ and $q\colon A''\to A$ such that $gq=\mathrm{Id}_{A''}$ and $pf=\mathrm{Id}_{A'}$. In particular, $T(p)\circ T(f) = T(\mathrm{Id}_{A'}) = \mathrm{Id}_{T(A')}$, so $T(f)$ has a left inverse; therefore, $T(f)$ must be one-to-one.
Similarly, $T(g)\circ T(q) = T(\mathrm{Id}_{A''}) = \mathrm{Id}_{T(A'')}$, so $T(g)$ has a right inverse; therefore, $T(g)$ is onto.
Well, $V_i$ is an arbitrary object of the putative exact sequence, so if you checked exactness there, that means you're done.
Or, that means you're done checking that the sequence of $V_i$ is an exact sequence, which is not quite what you set out to prove.
You set out to prove that such an exact sequence of $V_i$ "is the same as having a collection of short ones such that...", and yet you merely showed one way: if you have a collection of short exact sequences, you can splice them up.
Now you must show that given a long exact sequence you can chop it up in short exact sequences.
To see that, a diagram is worth a thousand words:
where you define $K_n=\operatorname{ker}(f_n)$.
The map $K_n\to A_n$ is the inclusion of the kernel of $f_n$. To get the map $A_{n+1} \to K_n$, observe that:
$K_n=\operatorname{ker}(f_n)=\operatorname{im}(f_{n+1})=\frac{A_{n+1}}{\operatorname{ker}(f_{n+1})}=\frac{A_{n+1}}{\operatorname{im}(f_{n+2})}=\operatorname{coker}(f_{n+2})$.
Hence, define the map $A_{n+1}\to K_n$ as the quotient map of $A_{n+1}$ by the image of $f_{n+2}$.
It is clear by definition that $0\to K_{n+1}\to A_{n+1}\to K_n\to 0$ are short exact sequences.
$ $
Now, you worry about how to chop a long exact sequence into short exact sequences. You shouldn't worry, since if
is an exact sequence, then you can enlarge it to an exact sequence infinite on both sides:
$ $
I wrote a couple of pages about splicing and decomposing exact sequences some time ago, which has a couple more results that you might find useful. There is also a section on equivalent definitions of exact functors which exploits these considerations. It is here: it's in Spanish, though.
Best Answer
In a general (abelian) category, the product of epimorphisms (if it exists!) may not be an epimorphism – that is why we sometimes assume (AB4*) as an extra axiom. As it turns out, the category of $R$-modules (for any ring $R$) always satisfies the (AB4*) axiom.
Proposition. If the (AB4*) axiom is satisfied in an abelian category, then products of short exact sequences are short exact sequences.
Proof. Consider short exact sequences $$0 \longrightarrow L^{(i)} \longrightarrow M^{(i)} \longrightarrow N^{(i)} \longrightarrow 0$$ By general abstract nonsense, one can show that the kernel of a product is the product of the kernels, so we have a left exact sequence $$0 \longrightarrow \prod_i L^{(i)} \longrightarrow \prod_i M^{(i)} \longrightarrow \prod_i N^{(i)}$$ but by the (AB4*) assumption, the last homomorphism is an epimorphism, so we in fact have a short exact sequence.