Let $F$ be a presheaf of abelian groups on $X$ (topological space for example, nice enough) and $\tilde{F}$ the associated sheaf. Then if $f\in H^0(X,\tilde{F})$, we can find a cover $\mathfrak{U}=(U_i)$ of $X$ and elements $f_i \in F(U_i)$ such that $f_i|_{U_i\cap U_j}\cong f_j|_{U_i\cap U_j}$ and $f|_{U_i} = f_i$.

Does the same hold if we omit abelian, i.e. if $F$ is just a sheaf of groups? Of course, in this case there are no cohomology groups but pointed sets. I guess it should but I don't know any reference.

  • $\begingroup$ My mistake. I understand you now. $\endgroup$ – Hoot Jun 30 '16 at 0:12
  • $\begingroup$ As I recall, Gunning's old notes on Riemann surfaces discussed (briefly) the general case. $\endgroup$ – Ted Shifrin Jun 30 '16 at 1:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.