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Is the left exactness of inverse limit (in the category of modules over a ring) a general property regardless of the indexing set? (Let's assume it is still directed.) The only proof I can find requires integers as the indexing set.

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The inverse limit is right adjoint to the diagonal functor, hence it is left exact because right adjoints preserve limits. –  t.b. Oct 18 '11 at 16:48
@t.b. Do you think there is a more elementary proof? –  ashpool Oct 18 '11 at 17:06
@ashpool: that is an elementary proof! –  Mariano Suárez-Alvarez Oct 18 '11 at 17:09
@ashpool: You can prove it "directly" (not "more elementary") using the construction. The only difficult part is showing exactness "in the middle", and you can prove that if $(m_i)$ lies in the kernel of $(g_i)$, then you can find a consistent family $(m'_i)$ in $\lim M'_i$ with $(f_i)(m'_i)=(m_i)$ by using Zorn's Lemma on a suitable family of "partial choices of $m'_i$. But using the universal properties via the adjoint functor is much simpler. –  Arturo Magidin Oct 18 '11 at 17:16
@ashpool: Actually, you don't even need Zorn's Lemma: the inverse limit is a submodule of the product, and the product construction is exact. You just need to show that if $(g_i)(m_i) = (0)$ then there is an $(m'_i)$ with $(f_i)(m'_i) = (m_i)$. You know it exists in the product, so you just need to show it is consistent. Use the fact that you have commuting squares and all the $f_i$ are one-to-one to show that this family must be consistent, hence lie in $\lim M'_i$. –  Arturo Magidin Oct 18 '11 at 17:27
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up vote 4 down vote accepted

The limit is left exact in that category, regardless of the indexing set.

If you think of the limit of a family as the submodule of the product subject to the compatibility conditions, then the induced morphisms between the limits are just the restrictions of the product functions. If $0\to M'_i\stackrel{f_i}{\to} M_i \stackrel{g_i}{\to} M''_i \to 0$ is exact, then each $f_i$ is one-to-one, so the induced map $$\lim M'_i \quad\stackrel{(f_i)}{\longrightarrow }\quad \lim M_i$$ is one-to-one, because $(f_i)$ is just the restriction of the product map $\prod M'_i\to \prod M_i$ to the submodule given by the limit, and this is the restriction of a one-to-one map (each component is one-to-one, so the product map is one-to-one).

Likewise, the composition $(g_i)\circ(f_i)$ is the zero map, because at each component you get $g_i\circ f_i = 0$; since each component of the map is the zero map, the composition is the zero map.

Finally, if $(m_i)$ lies in the kernel of $(g_i)$, then $m_i\in\mathrm{ker}(g_i)$ for each $i$, hence for each $i$ there is an $m'_i\in M'_i$ with $m_i = f_i(m_i)$; since each $f_i$ is one-to-one, $m'_i$ is uniquely determined. The difficulty lies in showing that this element $(m'_i)$ lies in $\lim M'_i$. But if $j,k\in I$ with $j\leq k$, then we know that $p_{kj}(f_j(m'_j)) = p_{kj}(m_j) = m_k = f_k(p'_{kj}(m'_j))$. Since $f_k$ is one-to-one, that means that $p'_{kj}(m'_j) = m'_k$ (as $f_k(m'_k)=m_k$), so the family is consistent.

If your limit is over a functor (instead of a partially ordered set) the same argument works, considering each arrow instead of each pair.

For more general categories the result still holds if the notion of "exactness" can be made to make sense; but the simplest way of doing it then is to invoke the fact that $\lim$ is a right adjoint, and therefore respect all limits, including "kernel" (equalizer of the identity and the zero map).

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