Bourbaki

Exercise 4, § 1

Algebra, I, 124

(a) Given a set E, let Φ\Phi be the set of mappings into E of any subset of E; if f,g,hf, g, h are three elements of Φ\Phi, show that if the composition (f∘g)∘h(f \circ g) \circ h is defined so is f∘(g∘h)f \circ (g \circ h), but that the converse is not true; if these two compositions are defined they are equal.

(b) Let F\mathscr{F} be a family of non-empty subsets of E such that no two have an element in common and let Ψ\Psi be the subset of Φ\Phi consisting of the bijective mappings of a set of F\mathscr{F} onto a set of F\mathscr{F}. Show that, under the law induced by the law f∘gf \circ g on Ψ\Psi, the condition of Exercise 3(a) holds.

Solution

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Transcribed from the printed volume. Not translated.