Proving the Invertibility of Composed One-to-One Functions

[John O' Meara]

Homework Statement

Prove: if f and g are one-to-one (i.e., invertible), then so is the composition $$f \circ g$$

Homework Equations

I think you prove that the composition f o g has an inverse? As, a function has an inverse if and only if it is one-to-one.

The Attempt at a Solution

$$(f \circ g)^{-1}((f \circ g)(x))=x\\ (f \circ g)((f \circ g)^{-1}(x))=x. \\ (f \circ g)^{-1}((f \circ g)(x))= (f(g(x)))^{-1}(f(g(x))) = f^{-1}(g(x)^{-1})(f(g(x)))$$. I wonder can I do just what I have just done? Help gratefully received, thank you.

 

[JG89]

The Attempt at a Solution

$$(f \circ g)^{-1}((f \circ g)(x))=x\\ (f \circ g)((f \circ g)^{-1}(x))=x. \\ (f \circ g)^{-1}((f \circ g)(x))= (f(g(x)))^{-1}(f(g(x))) = f^{-1}(g(x)^{-1})(f(g(x)))$$. I wonder can I do just what I have just done? Help gratefully received, thank you.

You can't assume f(g(x)) is invertible, because just as you said, a function is invertible if and only if it is one to one, so your assumption is equivalent to what you are trying to prove.

Remember that g(x) is one to one if and only if g(a) = g(b) implies a = b, for any a, b in the domain of g.

Can you prove that f(g(x)) is one to one now?

 

[John O' Meara]

I didn't know that "g(x) is one to one if and only if g(a)=g(b), etc,". I know that if (a,b)is a point on the graph of y=g(x) then b=g(a), which is equivalent to the statement that $$a=g^{-1}(b)$$, which means that (b,a) is a point on the graph of g_inverse. And that then g and g_inverse are symmetrical about the line y=x.

 

[JG89]

Yes but you're trying to prove that f(g(x)) is one-to-one.

Our function g(x) is invertible if and only g(a) = g(b) implies a = b. That is one of the definitions of one-to-one.

f is also one to one. So by the definition of one-to-one we can say...

 

[John O' Meara]

We can say that g and f have got inverse functions in consequence of being one to one, which states that: a function that assigns distinct outputs to distinct inputs is one to one.

 

[John O' Meara]

So we can write $$f^{-1}(f(x))=x. f(f^{-1}(x))=x.$$ And the same is true for g. But I still cannot see how you can write f o g is one to one?

 

[snipez90]

There is no need to apply inverses. JG89's definition of one-to-one is probably the most commonly utilized one, and it is clearly the easiest to apply here. The contrapositive simply states if you have two distinct inputs (a =/= b), then the outputs are distinct (f(a) =/= f(b)). An equivalent definition is that for each point b in the image of the domain of the function, there is only one element a in the domain such that f(a) = b. But all you have to prove is that for a and b in the domain of f composed with g, if f(g(a)) = f(g(b)) then a = b.

 

[Дьявол]

Take an example:

$$f(x)=\frac{1}{x}$$

f(x) is invertible i.e:

$$y=\frac{1}{x}$$

$$x=\frac{1}{y}$$

If the inverse is g(y) or

$$g(y)=\frac{1}{y}$$

we can write it as:

$$g(x)=\frac{1}{x}$$

so that f(x)=g(x), and the composition f(g(x))=f^-1(x)=g^-1(x)

Regards.

 

[John O' Meara]

Thanks for the replies, it looks very easy now that it is done.