Exercise 6, § 3 Tag 03I0, permanent. Cite it as [Bourbaki, Tag 03I0]. Theory of Sets, II, 124 For each graph G, let G′ be the graph ((pr1G)×(pr2G))−G((\mathrm{pr}_1\mathrm{G}) \times (\mathrm{pr}_2\mathrm{G})) - \mathrm{G}((pr1G)×(pr2G))−G. Show that (G−1)′=G′‾−1,(\overset{-1}{\mathrm{G}})' = \overset{-1}{\overline{\mathrm{G}'}},(G−1)′=G′−1, and that G∘(G−1)′⊂ΔB′,(G−1)′∘G⊂ΔA′\mathrm{G} \circ (\overset{-1}{\mathrm{G}})' \subset \Delta'_{\mathrm{B}}, \qquad (\overset{-1}{\mathrm{G}})' \circ \mathrm{G} \subset \Delta'_{\mathrm{A}}G∘(G−1)′⊂ΔB′,(G−1)′∘G⊂ΔA′ if A⊃pr1G\mathrm{A} \supset \mathrm{pr}_1\mathrm{G}A⊃pr1G and B⊃pr2G\mathrm{B} \supset \mathrm{pr}_2\mathrm{G}B⊃pr2G. Show that G=(pr1G)×(pr2G)\mathrm{G} = (\mathrm{pr}_1\mathrm{G}) \times (\mathrm{pr}_2\mathrm{G})G=(pr1G)×(pr2G) if and only if G∘G′−1∘G=∅.\mathrm{G} \circ \overset{-1}{\mathrm{G}'} \circ \mathrm{G} = \emptyset.G∘G′−1∘G=∅. Solution No solution has been written yet. Cited by Nothing. Cites Nothing.