Roos’ Theorem – Identifying the Error in the Proof

abelian-categoriesag.algebraic-geometryat.algebraic-topologyct.category-theoryhomological-algebra

Background: In 1961, Roos (who, sadly, apparently passed away just last month) purported to prove [1] that in an abelian category with exact countable products (AB4${}^\ast_\omega$), limits of inverse systems of epimorphisms are exact [2] [3]. The result stood for 41 years before Neeman published a counterexample in 2002. Influenced by Gabber, Roos published a corrected statement a few years later: in an abelian category with arbitrary coproducts (AB3) and exact countable products (AB4$^\ast_\omega$) which contains a generator, limits of inverse systems of epimorphisms are exact [4].

Question:
In neither Neeman's paper nor in Roos' 2006 paper can I find a discussion of where in his 1961 paper Roos actually makes a mistake. Can anybody shed any light on this?


Notes:

[1] Roos, Jan-Erik (1961). "Sur les foncteurs dérivés de lim. Applications". C. R. Acad. Sci. Paris. 252: 3702–3704. MR 0132091.

[2] By an "inverse system" $X^\bullet$, I mean a diagram $\dots \to X^n \to X^{n-1} \to \dots$ indexed by $\mathbb N$ (or $\mathbb Z$ — it doesn't make a difference). By an "inverse system of epimorphisms" $X^\bullet$, I mean an inverse system where each map $X^n \twoheadrightarrow X^{n-1}$ is an epimorphism. In an abelian category $\mathcal C$ with countable products (AB3$^\ast_\omega$), there is a limit functor from inverse systems in $\mathcal C$, to $\mathcal C$, which is left exact. If the countable products in $\mathcal C$ are exact (AB4$^\ast_\omega$), then there is a natural two-term complex $\prod_n X^n \to \prod_n X^n$ which may be used to derive the limit functor (the resulting $\delta$-functor is effaceable via the map $X^\bullet \to \prod_{n \leq \bullet} X^n$ where the latter complex uses the projections for structure maps); because this complex has two terms, the derived functors $\varprojlim^d X^\bullet$ vanish for $d > 1$. When I say that $X^\bullet$ is exact, I mean that $\varprojlim^1 X^\bullet = 0$. I think I understand how to argue that if $X^\bullet$ is an epimorphic system in an AB4$^\ast_\omega$ category and $\varprojlim^1 X^\bullet = 0$, then the map $\varprojlim X^\bullet \twoheadrightarrow X$ is an epimorphism for each $n$ (actually it's the reverse implication that holds); when Roos' theorem holds, this is a formulation of the result which doesn't mention $\varprojlim^1$.

[3] There is a stronger statement about Mittag-Leffler sequences; let me stick with the statement about inverse systems of epimorphisms for simplicity.

[4] In the comments here Leonid Positselski points out that Roos' proof actually establishes that this holds more generally in any AB4$^\ast_\omega$ category with enough projective effacements (I hadn't heard of this concept until recently, but it's right there in Tohoku — actually, enough local projective effacements suffices). In this form, the statement is actually quite straightforward to prove.

Best Answer

Taking a brief look at the Roos note we see that detailed proofs of the statements aren't provided. There is no argument in the note one could say is wrong. The paper with the corrected statement very carefully points out which parts of the original note do hold and which ones have to be modified and how and gives detailed arguments or precise references for the proofs of various results. See discussion at the start of Section 1 and at the start of Section 2. Thus, if you wanted to use this kind of result in another paper I would reference the newer paper.

In your question in your second footnote you have a sketch of an argument of something. But it suffers the same fate as the note of Roos of not being sufficiently written out to be sure if you have understood all possible pitfalls. I suggest you just write out completely what you were trying to say: the mathematics will guide you to the correct proofs and statements. The result should be longer than a Mathoverflow post; if not then you probably have not written it out completely. Good luck!

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