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By Daniel Bismuth Slavoj Zizek

242pages. in8. Broché.

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Example text

On note X ~1 le sous-espace de ~ form4 des fonctions correctes qui ont exactement deux v~leurs critiques 6gales. Ona: aln =¢. ifaut et il suffit Proposition 3'. Pour qu,un chemin dans qu'il soit correct, ~ valeurs dans nombre fini de valeurs de ~o 0 ~I, ~u'il ne rencontre ~1 que pour u_~n , e_~tque, pour chacune de ces valeurs, les deux branches du graphique se croisent transversalement. Remarque. Un chemin correct rencontre min excellent rencontre $~I Proposition ~ 1 e n u n n o m b r e fini de points ;donc tuu che- e n u n n o m b r e fini de points.

On suppose que et on note (Xo ' Yo ' Xo) = O. D2(O ) ~ O, localement en fonction du param&tre z = Les notations r , s , t , x o = Yo = X = O, o On note : D(p , q) ~(y , ~) = DI C o ~ e an 2 ° , on suppose (D'apr~s ,(p, D(~, ; q) x) = D2 " ce qui e n t r a i n e u f(~(u) du 2 ° , , y(u) ~, ~ , ~ , ? r(O) % 0 ; le graphique e s t d 4 f i n i par les 6quations : , ~k(u)) qu'on va utiliser sent celles d4finies par - les dquations 20 - dp dq = O. 6 (10) dl~ d2z dD1 dD2 d6 d~2 ~ ~ +Pd-~-+q ~-~-+ ~ D'oh aZ(o~ =0 du ; ~(o) = ~(0) d6 au _- + ~(o)~(o) ,_ du 3 du D o n e , c o m p t e t e n u de (9) az~(o)a3~(o~ du 2 : a6 ~(o) du 3" ' d26(o) du2 2 ~2(o)~(o) , P(O) d6(o" 2d~ d6 + ~(o) d26(0) ) a-d-(o)~(o) au2 u du et cette dernibre quamtitd est non nulle car | du = D(x , y , k) = et par cons6quent s : D~(o) ~(o)=-~o On peut 4noncer : s_~ l'indicatrice, r .

Bole ; o chemin A--~(f~ En plus, o_~n,cut choisir -__si fo E ~ 1 : , A) soi ~ e x c e l l e n t . u% c o r r e s p o n d e n t r e s p e c t i v e m e n t alors 6gales d~e f o , ~l(f~ ' X) - ~2(f~ , 4) (par c o n t i n u i t 4 ) auxvaleurs change d_~esigne en m~me temps (~ - ½)° ~o Remarque. On peut interpr4ter la proposition 6 en disant que ~o U ~ I une "subdivision cocellulaire" de et ~1 de codimension pour laquelle ~o O, d4finissent est de codimension 0 f o £ ~1 ( l e cas oh est plus simple).

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