M25008PP4

# M25008PP4 - Mathematics 250 Fall 2008 Proof Practice 4...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the 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

{[ snackBarMessage ]}

### Page1 / 2

M25008PP4 - Mathematics 250 Fall 2008 Proof Practice 4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online