Lemma 5
Tag 021K, permanent. Cite it as [Bourbaki, Tag 021K].
Topologie algébrique, chapter IV, § 2. Groupes de Poincaré des espaces délaçables
The set Γ is a meagre subset of .
Let Z denote the space . The topology of the space is defined by the distance given by sup and it is complete for this distance (TG, X, p. 20, corollaire and TG, X, p. 9, cor. 1); after TG, X, p. 24, th. 1, this topology is of countable type. The space is therefore a Polish space. The same is true of the space Z, for K is a Polish space (TG, IX, p. 57, prop. 1).
Let H be the subset of Z consisting of the triples such that is a strict homotopy joining to . The mappings from Z into given by and are continuous, for every , because the mapping from K into is continuous, and analogously the mappings from into X.
By definition, H is the intersection of the sets and , for , where denotes the diagonal of X. This proves that H is a closed subset of Z.
It follows that H is a Polish space. Let be the canonical projection; we have by definition. Since is separated, Γ is a Souslin subset of (TG, IX, p. 59, déf. 2). After IV, p. 362, prop. 7, it is an approachable subset of .
Suppose that Γ is not meagre. The space is a compact topological space and therefore a Baire space (TG, IX, p. 55, th. 1). Endow the group of permutations of the set with the discrete topology; let the product topological group operate diagonally in . The group G then operates continuously in by the mapping . After Lemma 4, the set V of elements such that is a neighbourhood of the identity element of G.
Let ; let . Since , we have ; since , so that . We have thus proved that, for every , there exists such that and are equivalent for R.
For , let denote the element of G all of whose terms are equal to except that with index , which is equal to the nontrivial element of . There exists an integer such that belongs to V; let then be such that and are equivalent for R. This implies that the loops and are strictly homotopic. By construction, these loops coincide on the intervals and ; on the interval , one is the constant mapping with image and the other is the mapping . It follows that is strictly homotopic to the constant loop with image , hence a contradiction. Lemma 5 is thus proved.
Let us now complete the proof of theorem 2. By Prop. 8, there exists a continuous injective mapping of into K whose image meets each equivalence class for R in at most one point. If and are distinct elements of , the loops and are not strictly homotopic in X, and the mapping given by is injective. In particular, Card() Card() Card(). Since X is a metrizable topological space of countable type, the same is true of (TG, X, p. 24, th. 1). Hence is homeomorphic to a subspace of (TG, IX, p. 18, prop. 12) and
Card(Ω) Card([0) Card()
= Card() Card().
A fortiori, Card() Card(). It then follows from Cor. 2 of E, III, p. 25, that has the power of the continuum, which was to be proved.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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