Bourbaki

Example

Tag 0DRO, permanent. Cite it as [Bourbaki, Tag 0DRO].

Algebra, chapter X, § 2. Complexes de A-modules

Consider the product of spheres S2×S2S_2 \times S_2 and the complex projective space P2(C)P_2(\mathbf{C}). Let b∈S2b \in S_2, we define a cellular decomposition (Yn)(Y_n) of Y=S2×S2Y = S_2 \times S_2 by setting

Y0=Y1={(b,b)},Y2=Y3=({b}×S2)∪(S2×{b})andY4=S2×S2; Y_0 = Y_1 = \{ (b, b) \} , \quad Y_2 = Y_3 = (\{ b \} \times S_2) \cup (S_2 \times \{ b \}) \quad \text{and} \quad Y_4 = S_2 \times S_2 ;

this decomposition has one cell of dimension 0, two of dimension 2 and one of dimension 4. The differentials of the associated complex are necessarily zero, so H0(Y)H_0(Y), H2(Y)H_2(Y) and H4(Y)H_4(Y) are free of rank 1, 2 and 1 respectively, and Hn(Y)=0H_n(Y) = 0 for n∉{0,2,4}n \notin \{ 0, 2, 4 \}. One obtains a cellular decomposition (Zn)(Z_n) of P2(C)P_2(\mathbf{C}) by setting

Z0=Z1={c},Z2=Z3=P1(C),Z4=P2(C), Z_0 = Z_1 = \{ c \} , \quad Z_2 = Z_3 = P_1(\mathbf{C}) , \quad Z_4 = P_2(\mathbf{C}) ,

the space P1(C)P_1(\mathbf{C}) being embedded in P2(C)P_2(\mathbf{C}) (TG, VIII, p. 20), and c being a point of P1(C)P_1(\mathbf{C}); this decomposition has one cell of dimension 0, one of dimension 2 and one of dimension 4. Here again the differentials of the complex are necessarily zero, and it follows that Hn(P2(C))H_n(P_2(\mathbf{C})) is isomorphic to A for n∈{0,2,4}n \in \{ 0, 2, 4 \} and to 0 otherwise. Since the homology modules in degree 2 of the two spaces considered are free of rank 2 and 1 respectively, these spaces are not homeomorphic. \*

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Nothing.

Translated with machine help, not yet reviewed by a person.