- #1
calvino
- 108
- 0
Show that there exists a mapping from a set S to itself that is 1-1, but not onto IFF there exists a mapping from a set S to itself that is onto, but not 1-1.
Firstly i show it, assuming that the one-one mapping (but not onto) exists.
Now i know that if there is a funtion that is 1-1, but not onto, to define a mapping f on the set (call it S), to a subset of S (call it T). This is such that f: S -> T is one-one and onto. Next, I define a mapping g:S->S such that
g(x) = { f^-1(x), when x is an element of T
{ x, whenever x is an element of S\T
That is my mapping that is then onto, but not one-one.
However, I'm stuch on the converse. That is, how do i prove, assuming a mapping from S to itself that is onto, but not one-one, implies that there is a mapping that is one-one, but not onto. ANY HELP>?
Firstly i show it, assuming that the one-one mapping (but not onto) exists.
Now i know that if there is a funtion that is 1-1, but not onto, to define a mapping f on the set (call it S), to a subset of S (call it T). This is such that f: S -> T is one-one and onto. Next, I define a mapping g:S->S such that
g(x) = { f^-1(x), when x is an element of T
{ x, whenever x is an element of S\T
That is my mapping that is then onto, but not one-one.
However, I'm stuch on the converse. That is, how do i prove, assuming a mapping from S to itself that is onto, but not one-one, implies that there is a mapping that is one-one, but not onto. ANY HELP>?
