Bourbaki

Exercise 29, § 11

General Topology, I

Bourbaki marks this one as harder than the rest.

Let YY be a locally compact, connected and locally connected space, NN a closed subset of YY such that, for every connected nonempty open subset VV in Y,V∩(Y−N)Y, V \cap (Y - N) are nonempty and connected. Let XX be a separated space, f:X→Y−Nf : X \to Y - N a proper local homeomorphism (exerc. 27). (a) Now prove that there exists a locally compact, connected and locally connected space ZZ, a homeomorphism jj of XX onto an everywhere dense open subset of ZZ and a proper mapping gg of ZZ into YY, such that g(y)g(y) is finite for every y∈Yy \in Y and that f=g∘jf = g \circ j. (First observe that for every connected open set W⊂Y−NW \subset Y - N, and every connected component CC of f−1(W)f^{-1}(W), one has f(C)=Wf(C) = W; consequently, if nn is the cardinal of f−1(y)f^{-1}(y) for every y∈Y−Ny \in Y - N, −1f(W)^{-1}f(W) has at most nn connected components. Deduce that for every y∈Ny \in \mathbf{N}, there are at most filters on X\mathbf{X} having a basis formed of connected components of the sets −1f(V)^{-1}f(V), where V ranges over the set of connected open neighbourhoods of yy in Y\mathbf{Y}. Let S be the set of all these filters when yy ranges over N\mathbf{N}. Take for ZZ a sum set of X\mathbf{X} and S; the neighbourhoods of a point F∈S\mathfrak{F} \in S are defined as follows: F\mathfrak{F} admits a basis formed of connected components C of sets −1f(V)^{-1}f(V), where V ranges over the set of connected open neighbourhoods of a point y∈Ny \in \mathbf{N}. For each of these sets C, consider the set C~\tilde{C} of filters F′∈SF' \in S to which C also belongs, and the sets C~\tilde{C} thus defined form by definition a fundamental system of neighbourhoods of F\mathfrak{F} in ZZ).

(b) Now prove that if Z1Z_1, g1g_1 and j1j_1 are a locally compact connected and locally connected space and two mappings having the same properties as ZZ, gg and jj, there exists a homeomorphism hh of ZZ onto Z1Z_1 such that j1=h∘jj_1 = h \circ j and g=g1∘hg = g_1 \circ h. (Use exerc. 28).

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-6. Not checked by a person.