1) Lets say we have a function f:X \rightarrow Y that has an inverse function. How do I find the function I(x) = f^{-1}(f(x)) and how can I find the domain and range of I(x) ? This one is very confusing and I love an good explanation for it.

Raynor2i 2022-07-17 Answered
Discrete math question about surjective, injective function and domain, range
1) Lets say we have a function f : X Y that has an inverse function. How do I find the function I ( x ) = f 1 ( f ( x ) ) and how can I find the domain and range of I(x) ? This one is very confusing and I love an good explanation for it.
2) Prove that if f and g are both surjective, then g∘f is surjective. I think that I have to prove that its image is equal to its codomain, but I have no idea how to do this.
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Answers (2)

lelapem
Answered 2022-07-18 Author has 12 answers
Step 1
1) Saying that f : X Y has an inverse is equivalent to saying that f is bijective, otherwise f 1 : Y X couldn’t be defined. Furthermore, composing the two always gives the Identity function, whether X X or Y Y depends on the order of composition (which one is it, in your example?).
Step 2
2) To prove stuff like this it is always a good idea to work with a specific element of your set and prove the inclusions in question. So, looking back at the definition: f : X Y is surjective iff y Y , x X such that f ( x ) = y (which is the same thing you said). Now, pick a z in the domain of g f and try to prove its pre image x exists.
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Lillie Pittman
Answered 2022-07-19 Author has 4 answers
Step 1
That is not always easy to find the inverse function when you have general sets X and Y.
However, if X and Y are subsets of R , for instance, you can find f 1 with the following idea: Solve the equation f ( y ) = x for y, then y will give you the formula for the inverse.
For instance, consider the function f : ( 0 , 1 ) R given by f ( x ) = x 1 + x .
Then f is bijective. Let us find f 1 . Well:
f ( y ) = x y 1 + y = x y = x ( 1 + y ) y x y = x y ( 1 x ) = x y = x 1 x .
The inverse will be given by
f 1 ( x ) = x 1 x .
Step 2
Let us check how this works:
f ( f 1 ( x ) ) = f 1 ( x ) 1 + f 1 ( x ) = x / ( 1 x ) 1 + ( x / ( 1 x ) = x / ( 1 x ) ( 1 x + x ) / ( 1 x ) = x / ( 1 x ) 1 / ( 1 x ) = x .
Of the other hand:
f 1 ( f ( x ) ) = f ( x ) 1 f ( x ) = x / ( 1 + x ) 1 ( x / ( 1 + x ) ) = x / ( 1 + x ) ( 1 + x x ) / ( 1 + x ) = x / ( 1 + x ) 1 / ( 1 + x ) = x .
In general, if f : X Y is bijective then f 1 will have Y as domain, and X as image.
Finally, if f : X Y and g : Y Z are surjective then g f : X Z is surjective. In fact, given z Z there exists y Y such that g ( y ) = z since g is surjective. Once f is surjective, there exists x X such that f ( x ) = y. Hence, the element x X is such that
( g f ) ( x ) = g ( f ( x ) ) = g ( y ) = z ,
that is, every element of Z is in the imagem of g f, this show Z Im ( g f ). Since, Im ( g f ) is always a subset of Z, it follows Z = Im ( g f ), therefore g f is surjective.
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