determing if functions are inverses

[intothevoidx]

I was wondering if someone could explain to me the easiest method for determining if two functions are inverses of eachother?

 

 

Thanks

[Dave]

Presuming you have two bijections, [imath]f : A \to B[/imath] and [imath]g : B \to A[/imath], then one takes the composite of the functions. If [imath]f \circ g = \text{id}_B[/imath] and [imath]g \circ f = \text{id}_A[/imath] (where [imath]\text{id}_X[/imath] is the identify function [imath]\text{id}_X (x) = x[/imath]) then [math]g = f^{-1}[/math].

[the tree]

Is that, in less words:[math]g=f^{-1}[/math] iff [math]fg(x) = x[/math]

 

Also, what's with the [imath] tags? They don't seem to behave differently to the [math] tags, is this part of a new feature?

[Dave]

Not quite; it is necessary that [math]f(g(x)) = g(f(x)) = x[/math].

 

imath tags do display math a little differently (namely they typeset inline maths). For example, [math]\lim_{x\to\infty}[/math] vs. [imath]\lim_{x\to\infty}[/imath]

[shadowacct]

Yes, what Dave says is an important addition.

A nice counter example, where g ◦ f is an identity map, while f ◦ g is not an identity map:

 

Function f from R to RxR: f(x) = (x, x)

Function g from RxR to R: g(x, y) = x

 

(g ◦ f)(x) = g(f(x)) = g(x,x) = x

(f ◦ g)(x, y) = f(g(x, y)) = f(x) = (x, x) ≠ (x, y)

[Xerxes]

Not quite; it is necessary that [math]f(g(x)) = g(f(x)) = x[/math].

 

 

I'm confused. In your set-up you had [math]f: A\rightarrow B, g: B \rightarrow A[/math]. This implies [math] x \in A, f(x) \in B[/math]. If [math] g\circ f = id_A[/math] then [math]g(f(x)) = x[/math] as you say.

 

But if [math]f\circ g = id_B[/math] how can [math]f(g(x)) = x[/math]? This implies [math]x \in B[/math] also, and that A = B. Is this what you meant?

 

Wouldn't it be clearer if you said for all [math] x \in A, y \in B, g(f(x)) = x, f(g(y)) = y[/math]?

 

Certainly I see no requirement that in general [math]g(f(x)) = f(g(y))[/math] Or am I missing the point?

[Country Boy]

I'm confused. In your set-up you had [math]f: A\rightarrow B, g: B \rightarrow A[/math]. This implies [math] x \in A, f(x) \in B[/math]. If [math] g\circ f = id_A[/math] then [math]g(f(x)) = x[/math] as you say.

 

But if [math]f\circ g = id_B[/math] how can [math]f(g(x)) = x[/math]? This implies [math]x \in B[/math] also, and that A = B. Is this what you meant?

 

Wouldn't it be clearer if you said for all [math] x \in A, y \in B, g(f(x)) = x, f(g(y)) = y[/math]?

 

Certainly I see no requirement that in general [math]g(f(x)) = f(g(y))[/math] Or am I missing the point?

The two "x"s are different. In writing g(f(x))= x, yes, the assumption is that x is in the domain of f, the set A. In writing f(g(x))= x, the assumption is that x is in the domain of g, the set B.

 

It would have been more precise to say "For all x in A, g(f(x))= x and for all x in B, f(g(x))= x".

 

I, personally, would probably have said "For all x in A, g(f(x))= x and for all y in B, f(g(y))= y" to make the distinction even clearer.

[Xerxes]

Good. In which case, the statement that "it necessary that [math] f(g(x)) = g(f(x) = x [/math]" is ...ahem.. misleading.

[Dave]

Good. In which case, the statement that "it necessary that [math] f(g(x)) = g(f(x) = x [/math]" is ...ahem.. misleading.

 

Yes, it is. Let me clarify: I'm saying that it is a requirement that [imath]f \circ g = \text{id}_B[/imath] and [imath]g \circ f = \text{id}_A[/imath]. My desire to use the least amount of notation possible unfortunately caused some misunderstanding.

[Xerxes]

So, anyway. Returning to the OP.

I was wondering if someone could explain to me the easiest method for determining if two functions are inverses of each other?

 

First as has been said, more-or-less, if a function is bijective, you may assume it has an inverse. So what's a bijection? Here.

 

Let f: A --> B. If, for all z in B there exists x in A such that f(x) = z, the function f is said to be surjective.

 

If for some x and y in A, f(x) = f(y) in B implies x = y in A, f is said to be injective.

 

A function that is both injective and surjective is said to be a bijection.

 

How to "determine" bijectivity? One way would be to graph your function, i.e. x vs f(x). If any "line" passing through the +/- f(x) axis and parallel to the +/- x axis intersects the curve f(x) vs x at least once and no more than once, then you have your bijection, an invertible function.

 

Micky Mouse? Of course, hope it helps.

[Country Boy]

Okay, a bijective function HAS an inverse. But none of that addresses the question of how you determine whether or not two given functions ARE inverse to one another! Fortunately that has been pretty well handled already.