Bourbaki

Exercise 7, § 2

Tag 03WM, permanent. Cite it as [Bourbaki, Tag 03WM].

Algebraic Topology, I, p. 141

(a) Let X be the set [0; 1[. Let TX\mathscr{T}_X be the set of subsets U of X satisfying the two properties:

(i) The intersection U∩]0;1[U\cap ]0; 1[ is open ;

(ii) If 0∈U0\in U, there exists an integer p⩾1p\geqslant 1 such that 1/n∈U1/n\in U for every integer n⩾pn\geqslant p.

Exercise prove that TX\mathscr{T}_X is a topology on X.

(b) Let Z be the set [0,1[2×(Z/2Z)[0,1[^2\times (\mathbf{Z}/2\mathbf{Z}). Let TZ\mathscr{T}_Z be the set of subsets U of Z satisfying:

(i) The intersection of U with the complement of {(0,0)}×(Z/2Z)\{(0,0)\} \times (\mathbf{Z}/2\mathbf{Z}) is open in the space Z endowed with the product topology of X×X×(Z/2Z)X\times X\times (\mathbf{Z}/2\mathbf{Z});

(ii) For every a∈Z/2Za\in \mathbf{Z}/2\mathbf{Z} such that (0,0,a)∈U(0,0, a)\in U, there exists an integer p⩾1p\geqslant 1 such that, for every couple (m,n√)(m, n\surd ) of natural integers satisfying√\surd inf(m,nm, n)⩾p\geqslant p, one has (1/n,√1/m,a)∈√U(1/n,\surd 1/m, a)\in \surd U if (n−m2)(m−n2)>0(n-m2)(m-n2)>0, (1/n,1/m,a+1)∈U(1/n,1/m, a+ 1)\in U if (n−m2)(m−n2)<0(n-m2)(m-n2)<0, (0,1/m,a)∈U(0,1/m, a)\in U and (1/n,0,a)∈U(1/n,0, a)\in U.

Exercise prove that TZ\mathscr{T}_Z is a topology on Z.

(c) The sets X and Z are endowed with these topologies, and the space X×XX\times X with the product topology. Let p:Z→X×Xp: Z\rightarrow X\times X be the canonical projection. Exercise prove that pp makes Z a covering of degree 2 of X×XX\times X.

(d) Exercise prove that pp admits no continuous section.

(e) Let s:X×X→Zs: X\times X\rightarrow Z be the mapping given by (x,y)↦(x,y,0)(x, y)\mapsto (x, y,0). Exercise prove that for every x∈Xx\in X, the restriction of ss to X×{x}X\times \{x\} is continuous, analogously the restriction of ss to {x}×X\{x\} \times X is continuous.

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.