Connes–Karoubi long exact sequence for Fréchet sheaves
... Lemma 2.3. Let (X, OX ) denote the formal scheme obtained by completing an integral noetherian scheme X ′ of finite type over C along a closed primary integral subscheme X, as in example (a) above. Then, for each open subset U of X, the ring OX (U ) is an ultrametric Banach algebra, i.e., (X, OX ) d ...
... Lemma 2.3. Let (X, OX ) denote the formal scheme obtained by completing an integral noetherian scheme X ′ of finite type over C along a closed primary integral subscheme X, as in example (a) above. Then, for each open subset U of X, the ring OX (U ) is an ultrametric Banach algebra, i.e., (X, OX ) d ...
Topological pullback, covering spaces, and a triad
... A covering map p : E → Y is trivial provided there exists a discrete space D (possibly empty) and an isomorphism t : p → pr1 where pr1 : Y × D → Y . If Y is nonempty, then Y has infinitely many isomorphism classes of trivial covers, one for each cardinal number. Remarks 2.6. Covering maps, and morphi ...
... A covering map p : E → Y is trivial provided there exists a discrete space D (possibly empty) and an isomorphism t : p → pr1 where pr1 : Y × D → Y . If Y is nonempty, then Y has infinitely many isomorphism classes of trivial covers, one for each cardinal number. Remarks 2.6. Covering maps, and morphi ...
Connectedness and continuity in digital spaces with the Khalimsky
... Suppose that m ∈ U 0 . By the same argument also m − 1 ∈ U 0 so m − 1 ∈ U . This contradicts the fact that U and V are disjoint. Therefore m 6∈ U 0 . It follows that U 0 and V 0 are disjoint. But then U 0 and V 0 separate Z. This is a contradiction, since Z is connected. Further results in this di ...
... Suppose that m ∈ U 0 . By the same argument also m − 1 ∈ U 0 so m − 1 ∈ U . This contradicts the fact that U and V are disjoint. Therefore m 6∈ U 0 . It follows that U 0 and V 0 are disjoint. But then U 0 and V 0 separate Z. This is a contradiction, since Z is connected. Further results in this di ...