{"id":635,"date":"2021-08-07T06:00:07","date_gmt":"2021-08-07T04:00:07","guid":{"rendered":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/?p=635"},"modified":"2021-08-04T11:45:52","modified_gmt":"2021-08-04T09:45:52","slug":"una-funcion-tiene-inversa-por-la-derecha-si-y-solo-si-es-suprayectiva","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/una-funcion-tiene-inversa-por-la-derecha-si-y-solo-si-es-suprayectiva\/","title":{"rendered":"Una funci\u00f3n tiene inversa por la derecha si y solo si es suprayectiva"},"content":{"rendered":"<p>En Lean, que g es una inversa por la izquierda de f est\u00e1 definido por<\/p>\n<pre lang=\"text\">\n   left_inverse (g : \u03b2 \u2192 \u03b1) (f : \u03b1 \u2192 \u03b2) : Prop :=\n      \u2200 x, g (f x) = x\n<\/pre>\n<p>que g es una inversa por la derecha de f est\u00e1 definido por<\/p>\n<pre lang=\"text\">\n   right_inverse (g : \u03b2 \u2192 \u03b1) (f : \u03b1 \u2192 \u03b2) : Prop :=\n      left_inverse f g\n<\/pre>\n<p>y que f tenga inversa por la derecha est\u00e1 definido por<\/p>\n<pre lang=\"text\">\n   has_right_inverse (f : \u03b1 \u2192 \u03b2) : Prop :=\n      \u2203 g : \u03b2 \u2192 \u03b1, right_inverse g f\n<\/pre>\n<p>Finalmente, que f es suprayectiva est\u00e1 definido por<\/p>\n<pre lang=\"text\">\n   def surjective (f : \u03b1 \u2192 \u03b2) : Prop :=\n      \u2200 b, \u2203 a, f a = b\n<\/pre>\n<p>Demostrar que la funci\u00f3n f tiene inversa por la derecha si y solo si es suprayectiva.<\/p>\n<p>Para ello, completar la siguiente teor\u00eda de Lean:<\/p>\n<pre lang=\"lean\">\nimport tactic\nopen function classical\n\nvariables {\u03b1 \u03b2: Type*}\nvariable  {f : \u03b1 \u2192 \u03b2}\n\nexample : has_right_inverse f \u2194 surjective f :=\nsorry\n<\/pre>\n<p>[expand title=\u00bbSoluciones con Lean\u00bb]<\/p>\n<pre lang=\"lean\">\r\nimport tactic\r\nopen function classical\r\n\r\nvariables {\u03b1 \u03b2: Type*}\r\nvariable  {f : \u03b1 \u2192 \u03b2}\r\n\r\n-- 1\u00aa demostraci\u00f3n\r\nexample : has_right_inverse f \u2194 surjective f :=\r\nbegin\r\n  split,\r\n  { intros hf b,\r\n    cases hf with g hg,\r\n    use g b,\r\n    exact hg b, },\r\n  { intro hf,\r\n    let g := \u03bb y, some (hf y),\r\n    use g,\r\n    intro b,\r\n    apply some_spec (hf b), },\r\nend\r\n\r\n-- 2\u00aa demostraci\u00f3n\r\nexample : has_right_inverse f \u2194 surjective f :=\r\nsurjective_iff_has_right_inverse.symm\r\n<\/pre>\n<p>Se puede interactuar con la prueba anterior en <a href=\"https:\/\/leanprover-community.github.io\/lean-web-editor\/#url=https:\/\/raw.githubusercontent.com\/jaalonso\/Calculemus\/main\/src\/Una_funcion_tiene_inversa_por_la_derecha_si_y_solo_si_es_suprayectiva.lean\" rel=\"noopener noreferrer\" target=\"_blank\">esta sesi\u00f3n con Lean<\/a>.<\/p>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;lean&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n<p>[expand title=\u00bbSoluciones con Isabelle\/HOL\u00bb]<\/p>\n<pre lang=\"isar\">\r\ntheory Una_funcion_tiene_inversa_por_la_derecha_si_y_solo_si_es_suprayectiva\r\nimports Main\r\nbegin\r\n\r\ndefinition tiene_inversa_dcha :: \"('a \u21d2 'b) \u21d2 bool\" where\r\n  \"tiene_inversa_dcha f \u27f7 (\u2203g. \u2200y. f (g y) = y)\"\r\n\r\n(* 1\u00aa demostraci\u00f3n *)\r\nlemma\r\n  \"tiene_inversa_dcha f \u27f7 surj f\"\r\nproof (rule iffI)\r\n  assume hf : \"tiene_inversa_dcha f\"\r\n  show \"surj f\"\r\n  proof (unfold surj_def; intro allI)\r\n    fix y\r\n    obtain g where \"\u2200y. f (g y) = y\"\r\n      using hf tiene_inversa_dcha_def by auto\r\n    then have \"f (g y) = y\"\r\n      by (rule allE)\r\n    then have \"y = f (g y)\"\r\n      by (rule sym)\r\n    then show \"\u2203x. y = f x\"\r\n      by (rule exI)\r\n  qed\r\nnext\r\n  assume hf : \"surj f\"\r\n  show \"tiene_inversa_dcha f\"\r\n  proof (unfold tiene_inversa_dcha_def)\r\n    let ?g = \"\u03bby. SOME x. f x = y\"\r\n    have \"\u2200y. f (?g y) = y\"\r\n    proof (rule allI)\r\n      fix y\r\n      have \"\u2203x. f x = y\"\r\n        by (metis hf surjD)\r\n      then show \"f (?g y) = y\"\r\n        by (rule someI_ex)\r\n    qed\r\n  then show \"\u2203g. \u2200y. f (g y) = y\"\r\n    by auto\r\n  qed\r\nqed\r\n\r\n(* 2\u00aa demostraci\u00f3n *)\r\nlemma\r\n  \"tiene_inversa_dcha f \u27f7 surj f\"\r\nproof (rule iffI)\r\n  assume \"tiene_inversa_dcha f\"\r\n  then show \"surj f\"\r\n    using tiene_inversa_dcha_def surj_def\r\n    by metis\r\nnext\r\n  assume \"surj f\"\r\n  then show \"tiene_inversa_dcha f\"\r\n    by (metis surjD tiene_inversa_dcha_def)\r\nqed\r\n\r\nend\r\n<\/pre>\n<p>En los comentarios se pueden escribir otras soluciones, escribiendo el c\u00f3digo entre una l\u00ednea con &#60;pre lang=&quot;isar&quot;&#62; y otra con &#60;\/pre&#62;<br \/>\n[\/expand]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>En Lean, que g es una inversa por la izquierda de f est\u00e1 definido por left_inverse (g : \u03b2 \u2192 \u03b1) (f : \u03b1 \u2192 \u03b2) : Prop := \u2200 x, g (f x) = x que g es una inversa por la derecha de f est\u00e1 definido por right_inverse (g : \u03b2 \u2192 \u03b1) (f : \u03b1 \u2192 \u03b2) : Prop := left_inverse f g y que f tenga inversa por la derecha est\u00e1 definido por has_right_inverse (f : \u03b1 \u2192 \u03b2) : Prop := \u2203 g : \u03b2 \u2192 \u03b1, right_inverse g f Finalmente, que f es suprayectiva est\u00e1 definido por def surjective (f : \u03b1 \u2192 \u03b2) : Prop := \u2200 b, \u2203 a, f a = b Demostrar que la funci\u00f3n&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[17],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/635"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/comments?post=635"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/635\/revisions"}],"predecessor-version":[{"id":636,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/posts\/635\/revisions\/636"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/media?parent=635"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/categories?post=635"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/calculemus\/wp-json\/wp\/v2\/tags?post=635"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}