I have seen multiple definitions for sheaf cohomology and wanted to ask for the reason. One goes through injective resolutions and the other through flasque resolutions. For paracompact bases it holds (i think) injective $\rightarrow$ flasque $\rightarrow$ soft $\rightarrow$ acyclic. Wouldn't it be "best" to define cohomology for more cases (for acyclic resolutions)? Is it the existence of acyclic resolutions, thats usually not a given?
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$\begingroup$ Any topological space (or site) has enough injective sheaves of abelian groups, so as a definition this always works. Then one checks that acyclic resolutions suffice as well, so a posteriori you can use flabby resolutions (or soft ones if the space is sufficiently nice). $\endgroup$– R. van Dobben de BruynCommented Feb 6, 2023 at 15:36
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1$\begingroup$ The usual definition is by injective resolutions as that is a special case of a "derived functor" (Look in Lang's Algebra book, for instance). This viewpoint makes it possible to compare sheaf cohomology to other types of cohomology. The fact that flasque, soft, fine an acyclic sheaves also compute the cohomology are theorems. $\endgroup$– user473423Commented Feb 6, 2023 at 15:37
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2$\begingroup$ That a sheaf is acyclic is a statement about its cohomology. Thus defining cohomology in terms of it is circular. Injective sheaves have exactly the lifting property necessary to compare two resolutions. If you use flasque sheaves, you are hoping that they are acyclic, but it is nontrivial to prove that. This reminds me of FAC, where Serre uses Čech chomology, which is like hoping that intersections of affines are acyclic, but he struggles with it because it's nontrivial, because it requires the hypothesis that the scheme is separated. In both cases $H^1$ easily works. $\endgroup$– Ben WielandCommented Feb 6, 2023 at 19:32
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