The first derived functor of the inverse limit
Inverse limits are left exact but not exact, and the failure is measured by lim1, the first derived functor. It appears whenever a filtered object is completed, in the Milnor sequence relating the cohomology of a limit to the limit of cohomologies, and in the convergence of unbounded spectral sequences. The Mittag-Leffler condition — that the images in the inverse system stabilise — forces it to vanish and is the standard hypothesis in practice.
Learning objectives
- Explain why inverse limits fail to be exact.
- Define lim1 and compute it in a simple case.
- State the Mittag-Leffler condition.
- Identify the Milnor sequence and where lim1 appears.
Section 01Failure of exactness
Given a short exact sequence of inverse systems, the limit functor gives
The surjectivity of the third map can fail: a compatible family of elements of Cn need not lift to a compatible family in Bn, because each individual lift is possible but the choices may not be made consistently. lim1 records exactly that inconsistency.
For inverse systems indexed by the natural numbers, limn = 0 for n ≥ 2. So the sequence above is the complete story, which is why lim1 appears everywhere and lim2 never does in this setting.
Section 02The Mittag-Leffler condition
- Consider the inverse system … → A2 → A1 → A0.
- For each n, look at the images of Am in An as m increases.
- The system is Mittag-Leffler when these images stabilise for each n. Surjective transition maps are the commonest sufficient condition.
- Mittag-Leffler implies lim1 = 0.
- The converse fails in general, so the condition is sufficient but not necessary.
| System | lim1 |
|---|---|
| Surjective transition maps | 0 — Mittag-Leffler holds |
| ℤ →×p ℤ →×p … | The p-adic integers modulo ℤ — non-zero |
| Finite groups | 0 — images stabilise by finiteness |
| Filtration quotients of a bounded filtration | 0 — eventually constant |
Section 03Where it appears
Milnor sequence
For a space that is a colimit of subspaces, cohomology of the colimit sits in a short exact sequence with a lim1 of the cohomologies of the stages.
Spectral sequence convergence
An unbounded filtration converges to the intended target only when the relevant lim1 vanishes.
Completion
The homology of a completed complex differs from the limit of the homologies by exactly a lim1 term.
Ext of a direct limit
Ext against a colimit involves an inverse limit in one variable and hence a lim1 correction.
Profinite completion
Cohomology of a profinite group is a colimit over finite quotients; comparisons with the abstract group involve lim1.
Homotopy limits
The Bousfield–Kan spectral sequence has lim1 terms in its E2 page.
ReferenceFrequently asked questions
Why is there no lim<sup>2</sup> for sequences?
Because an inverse system indexed by the natural numbers has a two-term resolution by products, so the derived functors vanish above degree 1. For systems indexed by more complicated posets higher derived functors can be non-zero.
Is lim<sup>1</sup> ever computable explicitly?
For the standard examples yes — it is the cokernel of an explicit map between products. In general it is more often shown to vanish than computed, which is what the Mittag-Leffler condition is for.
Does the dual problem arise for direct limits?
No, in module categories: filtered colimits are exact, so there is no colim1. This asymmetry between limits and colimits is one of the most practically consequential facts in the subject.
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