Take , , and let be the skyscraper sheaf with value a nonzero abelian group at . Cover by and . The presheaf gives and . A nonzero section on and the zero section on agree on the overlap, whose skyscraper sections vanish, but cannot be glued on . Thus need not be a sheaf.
Sheafification preserves stalks. If , neighborhoods contained in are cofinal, so . If , no neighborhood of lies in , so every term defining the presheaf stalk is zero. Therefore
Solved by gpt-5.6-sol high.
On the defining presheaf, send a section over by the identity
and use the unique zero map when . These maps commute with restrictions and therefore sheafify to the natural counit
After restricting back to , every open set lies in , so one recovers the original sheaf . Equivalently, the natural map is an isomorphism on every stalk and hence an isomorphism of sheaves.
Solved by gpt-5.6-sol high.
Use the counit from part b for the first map and the restriction map for the second. Exactness can be checked on stalks. At the sequence is
whereas at it is
Thus
is a short exact sequence of sheaves.
Solved by gpt-5.6-sol high.
Let and . A global section of is a regular function on the integral scheme whose support is closed in and contained in . Every nonzero regular function on has support dense in , so
If this sheaf were quasi-coherent, then on the affine scheme it would be the sheaf associated with this zero module and hence would vanish. Its stalks at points of are instead by part a. This contradiction proves that extension by zero need not preserve quasi-coherence.
Solved by gpt-5.6-sol high.

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