How do we prove that a transformation is bijection and one to one function? thank you.

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- Dec 21st 2013, 03:35 AMdavidciprutTransformation
How do we prove that a transformation is bijection and one to one function? thank you.

- Dec 21st 2013, 03:54 AMromsekRe: Transformation
to prove one to one prove

$\displaystyle (f(x)=y) \wedge (f(z)=y) \longrightarrow x=z$

to prove surjection prove that

$\displaystyle \forall y \in Y\;\;\exists x \in X \ni y=f(x)$

$\displaystyle \mbox{where }Y\mbox{ and }X\mbox{ are the range and domain of f respectively.}$

bijection is surjection and one to one. - Dec 21st 2013, 07:58 AMHartlwRe: Transformation
- Dec 21st 2013, 12:16 PMDevenoRe: Transformation
By and large, it depends on what kind of tools we have on hand. If our transformation is LINEAR, for example, we have more options than if our transformation is just a function.

For any function (linear or not) injective means f(x) = f(y) implies x = y. For example, the function f: R-->R given by

f(x) = x+1 is injective:

If f(x) = f(y), then:

x+1 = y+1, so x = (x+1)-1 = (y+1)-1 = y.

If f is a linear transformation, it is often easier to prove one of the following:

a) the null space of f is the 0-vector only.

b) the set of images of a basis set is a linearly independent set.

Surjectivity is often "harder": if f has domain X and co-domain Y, we need to show that for any y in Y, there is SOME x in X with f(x) = y. For example, the function

f:R-->R, f(x) = x+1 is also surjective:

Let a be any real number. We have to find some real number x with f(x) = a. Clearly, a-1 will do:

f(a-1) = (a-1)+1 = a+(-1+1) = a+0 = a.

If f is a linear transformation it is often easier to prove one of the following:

a) rank(f) = the dimension of the co-domain

b) the set of images of a basis for the domain spans the co-domain

A bijective function is both injective and surjective. Such functions are said to be invertible. One can show bijectivity by proving a function is both surjective AND injective, or: by exhibiting an inverse function, that is, if:

f:X-->Y

a function g:Y-->X such that fg = identity on Y, and gf = identity on X.

For example, the function g:R-->R given by g(x) = x-1 is an inverse for our example f given above:

g(f(x)) = g(x+1) = (x+1)-1 = x+(1-1) = x+0 = x, so gf is the identity function on R, and:

f(g(x)) = f(x-1) = (x-1)+1 = x+(-1+1) = x+0 = x, so fg is also the identity function on R.

for a linear transformation T:U-->V it is often easier to prove:

a) the set of images of a basis for U is a basis for V, or

b) dim(T(U)) = dim(V) = dim(U).