This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Mathematics 250 Fall 2008 Proof Practice # 4 Suppose that D and E are open domains in R k , and that F : D → E is a function defined on D and taking values in E . We say that F is invertible iff there is a function G : E → D , so that G ◦ F = I D and F ◦ G = I E . Here I D is the identity map on D , and similarly for I E . We can also write these conditions as G ( F ( x )) = x for all x in D , and F ( G ( y )) = y for all y in E . The mapping G is called the inverse of F , and is usually denoted by F ?1 . Show that F : D → E is invertible if and only if i) F isone-to-one (which means that if x 1 6 = x 2 are two different points in D , then F ( x 1 ) 6 = F ( x 2 )). (One-to one is also called being injective .) ii) F is onto (which means that, for any y in E , there is z in D such that F ( z ) = y ). Hint: Your main job in this exercise is to define F- 1 . Remark: Notice that the definition of invertibility for F , while clear enough, is extrinsic , in the sense that it depends on finding another function, with different domain and range from...
View Full Document
- Fall '06
- Math, #, Inverse function, Lemma, different points