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Hi everybody,

I have a lack of references concerning projective limits and injective limits. Up to my faults in Bourbaki there are only proj and inj limits indexed by a partially ordered set (not a category), and this set is almost systematically directed (i.e. for all $i,j$ there exists $k$ such that $k\geq i$, and $k\geq j$). The problem of combining projective limits with inductive limits is not treated at all as far as I can see.

I need a reference for this. Can someone help me ? Thank you !

My specific problem is the following: Assume given a system of modules $(M_{i,j})_{i,j}$ and arrows among them, over the same ring $A$.

We consider both

1) the projective limit with respect to the index $j$ of the injective limits with respect to the indexes $i$. Call it

$\projlim_j \;\;(\injlim_i \;\; (M_{i,j}))$

2) the injective limit with respect to the index $i$ of the projective limits with respect to the indexes $j$. Call it

$\injlim_i \;\;(\projlim_j\;\; (M_{i,j}))$

-- (Up to errors) There exists a canonical map

$\displaystyle CAN : \injlim_i \;\;(\projlim_j\;\; (M_{i,j})) \;\to\; \projlim_j\;\; (\injlim_i\;\; (M_{i,j})) $

Under which assumptions this map is INJECTIVE ?

As an example if there is no arrows at all between the $M_{i,j}$, then one has a direct sum instead of the injective limit, and a product instead of a projective limit. The arrow CAN becomes

$\displaystyle CAN : \oplus_i \;\;\prod_j \;M_{i,j} \;\to\; \prod_j \;\;\oplus_i \;M_{i,j}$

where

$((a_{i,j})_i)_j \mapsto ((a_{i,j})_j)_i$

In this case the arrow CAN is always injective independently on the nature of the $M_{i,j}$ (and up to errors it is an isomorphism if and only if one of the sets of index "$i$" or "$j$" is finite). I suspect that in the general case the injectivity only depends on the nature of the arrows, but not of the nature of the objects. Does anyone have a useful comment or a reference?

Many thanks !

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The canonical morphism $\alpha : \mathrm{colim}_{i \in I} ~ \mathrm{lim}_{j \in J} M_{ij} \to \mathrm{lim}_{j \in J} ~ \mathrm{colim}_{i \in I} M_{i,j}$ is injective in the following situation:

1) $I$ is a directed set.

2) For every $j \in J$ and every $i \to i'$ in $I$ the morphism $M_{i,j} \to M_{i',j}$ is injective.

More detailed: Let us still assume 1) and let $x \in \mathrm{ker}(\alpha)$. Choose $i$ such that $x$ comes from an element in $\lim_j M_{i,j}$, say $x = (x_{ij})_j$. Now $\alpha(x)=0$ says that for all $j$ there is some $i(j) \geq i$ such that the image of $x_{ij}$ in $M_{i(j),j}$ vanishes. If we have 2), this already gives us $x_{ij}=0$ for all $j$, thus $x=0$.

More generally, if $j \mapsto i(j)$ is a bounded function, say by $i_{\infty}$, we see that the image of $x$ in $\lim_j M_{i_{\infty},j}$ vanishes, which implies $x=0$ in the colim-lim. Now it's easy to find an example where this function is not bounded and, in fact, $\alpha$ is not injective:

Let's take $I=\mathbb{N}$ as a partial order and $J = \mathbb{N}$ as a discrete category, so that we consider the canonical map $\mathrm{colim}_n \prod_m M_{nm} \to \prod_m \mathrm{colim}_n M_{n,m}$. For all $n,m$ let $M_{n,m} = A[X]$ and let $M_{n,m} \to M_{n+1,m}$ be the formal derivative of polynomials. Then $(1,X,X^2,X^3,\dotsc)$ represents a nontrivial element in the kernel of $\alpha$, since every $X^m$ satisfies $\partial^{m+1} X^m = 0$, but there is no $n$ with $\partial^{n} X^m = 0$ for all $m$.

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  • $\begingroup$ Thank you for the answer ! Do you have a counterexample to the general case ? $\endgroup$ Commented Feb 13, 2012 at 20:48
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    $\begingroup$ Have you tried to find one by yourself? $\endgroup$ Commented Feb 13, 2012 at 21:19
  • $\begingroup$ Yes, I have tried, but I think I am basically stupid. My "psycological problem" is that the two cases that we know (direct sum case, and the case that you found of a directed set with injective maps) seems to be somehow "orthogonal". What is the right assumption that should include both these two cases cases ? For this reason I am thinking that maybe the map CAN is always injective ? What is your feeling ? Thanks again $\endgroup$ Commented Feb 13, 2012 at 21:38
  • $\begingroup$ I've added an example and more details. $\endgroup$ Commented Feb 13, 2012 at 22:35

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