Left exactness of inverse limit 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.
 A: 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).
