Inverse Function Proof: f and g are Isomorphisms

[kathrynag]

Homework Statement

If f:x-->y has an inverse function g, then g:y--->x is one to one and onto

Homework Equations

The Attempt at a Solution

Let g be the inverse of f:x--->y
I think this must have something to do with an isomorphism because of one to one and onto.

 

[waht]

has more to do with the function being bijective which is both one-one and onto (injective and surjective)

just have to work the definitions, one-one means that only one distinct element gets mapped, and onto implies that the entire range has to be mapped, together these conditions guarantee that only one element gets uniquely mapped to another element throughout the domain we are working in.

 

[kathrynag]

Ok, I understand those defintions. My problem is understanding why g has to be bijective.

 

[e(ho0n3]

Do you know how to prove that a function is bijective?

 

[kathrynag]

show a(x)=a(y) implies x=y
onto, not entirely sure. I know it has something to do with showing entire range is used

 

[e(ho0n3]

onto, not entirely sure. I know it has something to do with showing entire range is used

Not the range, but the codomain, i.e. for every y in the codomain, there is an x in the domain with y = f(x).

 

[kathrynag]

Let g be the inverse of f:y--->x

Then g(a)=g(b)
a=b
Does this prove one to one? For some resaon I think it doesn't.

 

[e(ho0n3]

There is an intermediate step between "g(a) = g(b)" and "a = b". How would get rid of the g in g(a) to get a?

 

[kathrynag]

I'm blanking on that

 

[HallsofIvy]

Apply the inverse of g to both sides of the equation.

 

[kathrynag]

so g^(-1)g(a)=g^(-1)g(b)?

 

[e(ho0n3]

If g is the inverse of f, then what is the inverse of g? Also, what does g^{-1}(g(a)) evaluate to?

 

[kathrynag]

f is the inverse of g?

 

[e(ho0n3]

That's right. And what about the answer to my second question.

 

[kathrynag]

f(g(a))

 

[e(ho0n3]

And that simplifies to ...

 

[kathrynag]

a I think

 

[e(ho0n3]

That's right. Now what is the complete proof that g is one-to-one?

 

[kathrynag]

by showing a=b