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Using right derived functors of the global sections functor, I'd like to calculate the first cohomology group of the constant sheaf $\mathbf{Z}$ on $S^1$ with its usual topology, $H^1(S^1,\mathbf{Z})$. I understand this can be done using Čech cohomology, but I'd like to compute it with derived functors.

To do so, I need to write down an injective resolution of the constant sheaf $\mathbf{Z}$. One such injective resolution involves products of skyscraper sheaves, but this seems fairly messy if all I want to show is $H^1(S^1,\mathbf{Z})=\mathbb{Z}$. Is there a cleaner injective resolution of $\mathbf{Z}$ in $\mathfrak{Ab}(X)$, the category of sheaves of abelian groups on $X$?

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Injective resolutions are not easy things to find; that is why we have Čech cohomology! – Zhen Lin Apr 26 '13 at 6:42
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This is exervise 3.2.7 in Hartshorne's "Algebraic Geometry". A solution can be found on page 103 in this PDF: math.northwestern.edu/~jcutrone/Work/… – Fredrik Meyer Apr 26 '13 at 7:12
    
@FredrikMeyer: I've seen this solution, and it uses products of skyscraper sheaves. My question asks if there is a cleaner injective resolution. – Jared Apr 26 '13 at 13:32
    
@ZhenLin: Thank you for answering my other question, but I'm still trying to answer this one. In the linked answer given by Fredrik Meyer, the solution claims that $\prod i_*(\mathbb{Z})$ is an injective sheaf. Do they mean to replace $\mathbb{Z}$ with $\mathbb{Q}$? Also, would it follow that $0\to\mathbb{Z}\to\prod i_*(\mathbb{Q})\to\prod i_*(\mathbb{Q}/\mathbb{Z})\to 0$ is an injective resolution of the constant sheaf $\mathbb{Z}$? – Jared Aug 27 '14 at 20:52
    
I doubt $\prod i_* \mathbb{Z}$ is injective. I also don't think your sequence is exact – certainly not in general. – Zhen Lin Aug 27 '14 at 21:03

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