(* Imagen_de_imagen_inversa_de_aplicaciones_suprayectivas.thy -- Si f es suprayectiva, entonces u ⊆ f[f⁻¹[u]]. -- José A. Alonso Jiménez -- Sevilla, 11-junio-2021 -- ------------------------------------------------------------------ *) (* --------------------------------------------------------------------- -- Demostrar que si f es suprayectiva, entonces -- u \ f ` (f -` u) -- ------------------------------------------------------------------- *) theory Imagen_de_imagen_inversa_de_aplicaciones_suprayectivas imports Main begin (* 1\ demostración *) lemma assumes "surj f" shows "u \ f ` (f -` u)" proof (rule subsetI) fix y assume "y \ u" have "\x. y = f x" using \surj f\ by (rule surjD) then obtain x where "y = f x" by (rule exE) then have "f x \ u" using \y \ u\ by (rule subst) then have "x \ f -` u" by (simp only: vimage_eq) then have "f x \ f ` (f -` u)" by (rule imageI) with \y = f x\ show "y \ f ` (f -` u)" by (rule ssubst) qed (* 2\ demostración *) lemma assumes "surj f" shows "u \ f ` (f -` u)" proof fix y assume "y \ u" have "\x. y = f x" using \surj f\ by (rule surjD) then obtain x where "y = f x" by (rule exE) then have "f x \ u" using \y \ u\ by simp then have "x \ f -` u" by simp then have "f x \ f ` (f -` u)" by simp with \y = f x\ show "y \ f ` (f -` u)" by simp qed (* 3\ demostración *) lemma assumes "surj f" shows "u \ f ` (f -` u)" using assms by (simp only: surj_image_vimage_eq) (* 4\ demostración *) lemma assumes "surj f" shows "u \ f ` (f -` u)" using assms unfolding surj_def by auto (* 5\ demostración *) lemma assumes "surj f" shows "u \ f ` (f -` u)" using assms by auto end