Bourbaki

Exercise 28, § 5

General Topology, IX

Bourbaki marks this one as harder than the rest.

Let G,G′G, G' be two topological groups. Show that every continuous homomorphism ff of GG onto G′G' is a strict morphism in each of the following two cases:

a) GG is locally compact and σ\sigma-compact, and G′G' is non-meagre [show that there is a relatively compact open set UU in GG such that the interior of f(U‾)f(\overline{U}) is not empty; deduce that for each compact neighbourhood VV of the identity element of GG, the interior of f(V)f(V) is not empty, and then apply Chapter III, § 2, Exercise 18].

b) GG is complete, the topology of GG has a countable base, and G′G' is non-meagre. [By using the fact that, for each neighbourhood UU of the identity element ee of GG, GG is the union of a sequence of sets of the form xnUx_n U, show that for each open set AA in GG there is an open set A′A' in G′G' which contains f(A)f(A) and is such that f(A)f(A) is dense in A′A'. Then consider a fundamental system (Un)(U_n) of symmetric neighbourhoods of ee in GG, such that U‾n+12⊂Un\overline{U}_{n+1}^2 \subset U_n, and show that f(Up)f(U_p) contains the open set Up+1′U'_{p+1}; to do this, take any point a′∈Up+1′a' \in U'_{p+1} and then construct by induction a sequence (bn)(b_n) of points of GG such that bn∈Up+nb_n \in U_{p+n} and such that f(b1b2…bn)f(b_1 b_2 \ldots b_n) tends to a′a'; note that, if x′∈Uk′x' \in U'_k, then there exists y∈Uky \in U_k such that f(y)∈x′Uk+1′f(y) \in x'U_{k+1}'.] The hypothesis that GG has a countable base is essential, as is shown by the example where G′G' is the group R\mathbf{R} with the topology of the real line, GG is the group R\mathbf{R} with the discrete topology and f:G→G′f : G \to G' is the identity mapping.

Solution

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