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It is well-known that if you have an inverse system of abelian groups $(A_n)$ (this works in several other nice categories) in which all the maps are surjective (or at least satisfy the Mittag-Leffler Condition), and if you have a short exact sequence of inverse systems $0\to (A_n)\to (B_n)\to (C_n)\to 0$, then taking the limit is exact and you get another short exact sequence $0\to \lim A_n \to \lim B_n \to \lim C_n \to 0$.

Hartshorne warns that this is not the case with abelian sheaves on a space. In particular, you can have all the maps of $(\mathcal{F}_n)$ surjective, and a short exact sequence $0\to (\mathcal{F}_n)\to (\mathcal{G}_n)\to (\mathcal{J}_n)\to 0$ but you only get left exactness $0\to \lim \mathcal{F}_n\to \lim \mathcal{G}_n\to \lim \mathcal{J}_n$

I.e. you get that $\lim^1(\mathcal{F}_n)\neq 0$ despite satisfying surjectivity of maps. Is there a canonical example of this happening?

My first guess was that this had to be related to the fact that you can have a surjective map of sheaves $\mathcal{F}\to \mathcal{G}$, yet still have an open set for which $\mathcal{F}(U)\to \mathcal{G}(U)$ is not surjective. The canonical example of when this happens is to use the exponential map on the sheaf of holomorphic functions on $\mathbb{C}^\times$, but it is very non-obvious to me how to turn this into an example of the above.

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up vote 3 down vote accepted

Let $X = \text{Spec}\;ℂ[x]$, and $\mathcal{G}⊂\mathcal{O}_X$ be the subsheaf of regular (rational) functions with no pole at $0$ or $∞$, and at most simple poles elsewhere. Thus if $U = ℂ \backslash \lbrace a_1,\dotsc,a_m \rbrace$ or $ℂ \backslash \lbrace a_1,\dotsc,a_m,0 \rbrace$, with $a_1,\dotsc,a_m$ distinct and nonzero, then $\mathcal{G}(U) = \lbrace\dfrac{f}{∏_{i=1}^m(x-a_i)} \vert \deg(f)≤m \rbrace$. Then let $\mathcal{G}_n = \bigoplus_{d=1}^n\mathcal{G}$, with $\mathcal{G}_{n+1}↠\mathcal{G}_n$ by projection onto the first $n$ components. Let $\mathcal{F}_n = \bigoplus_{d=1}^n\mathcal{F}^d$, where $\mathcal{F}^d⊂\mathcal{G}$ is the subsheaf of functions with a zero of order at least $d$ at $0$. Then the sheaf (not presheaf!) $\mathcal{G} / \mathcal{F}^d$ is isomorphic to the constant sheaf $ℂ[x]/(x^d)$. Therefore for nonempty $U$, $(\varprojlim\;\mathcal{G}_n/\mathcal{F}_n)(U) = ∏_{d=1}^∞ ℂ[x]/(x^d)$, whereas $\varprojlim\;\mathcal{G}_n(U) = ∏_{d=1}^∞ \mathcal{G}(U)$. Passing to stalks, we see $\varprojlim\;\mathcal{G}_n → \varprojlim\;\mathcal{G}_n/\mathcal{F}_n$ is not surjective, since $ℂ(x) → ℂ[[x]]$ is not.

Remark: Obviously this example is not as simple or enlightening as yours. However, one can argue that the example needs to be at least this complex, I believe. Without going into details, let me at least say that surjectivity of a sheaf map is a local property, so it takes more than a global topological obstruction to block it.

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Thanks so much! I've figured a few things out (the sheaves can't be quasi-coherent) and have a few candidate examples (inspired from formal schemes) since I posted this, but I was still clueless about how to put anything together to get an example. I'm not sure I understand $\mathcal{G}$ based on the description over $U$. What is that degree condition? Shouldn't any regular function in the numerator work? I also don't see why $\mathcal{G}/\mathcal{F}^d$ is that constant sheaf, but I'll put more thought into that. – Matt Mar 31 '11 at 20:45
The idea is that working mod $x^d$, one can "rationalize denominators" - convert a rational function whose denominator has a nonzero constant term into a polynomial. However, for large $d$, this requires using high degree polynomials. Now one wants that for large $d$, sections of $\mathcal{G}$ over small open sets (large $m$) glue together mod $\mathcal{F}^d$, but not otherwise. Requiring no poles at $∞$ gives $\deg(f)≤m$, forcing $m$ large to make sections glue mod $x^d$. Hope this helps. – Dave Barton Apr 1 '11 at 0:41
Thanks. For some reason when I read this I pushed that condition about no poles at infinity out of my mind and ignored it. That's what caused my confusion. Now it makes sense. – Matt Apr 1 '11 at 18:01

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