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Undergrad; Bijectivity of a composite function

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Hi, i'm not looking for a direct solution, just a point in the right direction.


g∘f(x) = -2x4 +4x2 -1
Where I have g∘f : R (-∞,1]

I know that when a mapping is bijective, it has to be injective and surjective; and for it to be injective, each unique domain values map to a unique co-domain value, and for it also to be surjective, each value within the domain maps to any value within the range(image).

I have the range for g∘f to be (-∞,1], and because thats the same interval as the codomain, then the mapping is surjective.

For injection, I let g∘f(x) = 0 and solve the quadratic function by letting u = x^2
So i’m dealing with -2u^2 + 4u -1
So solving it i get u = 1 (√2)/2
Where u = x^2, x = √(1 (√2)/2)
As there are four unique x values for 1 domain value, then g∘f is not injective, meaning its not bijective.

Is this along the right lines? Or is there something big/fundamental that i’m missing here.

Thank you :smile:
(edited 5 years ago)
Original post by tom09099


Hi, i'm not looking for a direct solution, just a point in the right direction.


g∘f(x) = -2x4 +4x2 -1
Where I have g∘f : R (-∞,1]


Okay, first of all, this is not quite correct.

gf=g(f(x))g \circ f = g(f(x)). We know the input of gg is R\mathbb{R} and the output is (,1](-\infty, 1], which means we are allowed to have f(x)Rf(x) \in \mathbb{R} since that's our input. By how ff is defined, this means x(,1]x \in (-\infty, -1].

Hence the mapping gfg \circ f is (,1](,1](-\infty, -1] \to (-\infty, 1].

I know that when a mapping is bijective, it has to be injective and surjective; and for it to be injective, each unique domain values map to a unique co-domain value, and for it also to be surjective, each value within the domain maps to any value within the range(image).

I have the range for g∘f to be (-∞,1], and because thats the same interval as the codomain, then the mapping is surjective.

For injection, I let g∘f(x) = 0 and solve the quadratic function by letting u = x^2
So i’m dealing with -2u^2 + 4u -1
So solving it i get u = 1 (√2)/2
Where u = x^2, x = √(1 (√2)/2)
As there are two unique x values for 1 domain value, then g∘f is not injective, meaning its not bijective.

Is this along the right lines? Or is there something big/fundamental that i’m missing here.

Thank you :smile:


It is indeed surjective, but the reasoning doesn't hold justice for me.

We know that the codomain is (,1](-\infty, 1] so what you want to show is that your expression y=4x22x41y = 4x^2 - 2x^4 - 1 has range precisely <y1-\infty < y \leq 1 [HINT: Complete the square, it's just a disguised quadratic!]

That way, the range and the codomain coincide hence the function is surjective.


To show that the mapping fg:(,1](,1]f \circ g : (-\infty, -1] \to (-\infty, 1] is injective with f(x)=4x22x41f(x) = 4x^2-2x^4-1, it is probably easier to show the implication:

F(x1)=F(x2)    x1=x2F(x_1) = F(x_2) \implies x_1 = x_2

So 4x122x141=4x222x241    2(x12x22)=x14x244x_1^2 -2x_1^4 -1 = 4x_2^2 - 2x_2^4 - 1 \iff 2(x_1^2-x_2^2) = x_1^4 - x_2^4

Finish it off from there using the facts that x1,x21    x12,x221x_1, x_2 \leq -1 \implies x_1^2, x_2^2 \geq 1
(edited 5 years ago)
Original post by RDKGames

So since x1,x21x_1, x_2 \leq -1 it should be obvious that x1x2    x12x22x_1 \neq x_2 \implies x_1^2 \neq x_2^2 hence 4x124x224x_1^2 \neq 4x_2^2.

Similarly, since x12,x221x_1^2, x_2^2 \geq 1, we must have that x12x22    x14x24x_1^2 \neq x_2^2 \implies x_1^4 \neq x_2^4.

Finish it off from there.


:holmes: Not sure how you're intending the OP to finish off from there - it's not totally trivial since you'll be taking the difference of the two terms.

Not looking for an explanation - just checking you're happy with it.

Edit: Refers to earlier version of last post. This one can be ignored - retained for integrity of the thread.
(edited 5 years ago)
Original post by ghostwalker
:holmes: Not sure how you're intending the OP to finish off from there - it's not totally trivial since you'll be taking the difference of the two terms.

Not looking for an explanation - just checking you're happy with it.


I'm happy with it, but is there something I'm overlooking that doesn't add up for you?

Since x14x24x_1^4 \neq x_2^4 then x14x24-x_1^4 \neq -x_2^4 so it's just adding the two terms which preserves the inequality.
Original post by RDKGames
I'm happy with it, but is there something I'm overlooking that doesn't add up for you?

Since x14x24x_1^4 \neq x_2^4 then x14x24-x_1^4 \neq -x_2^4 so it's just adding the two terms which preserves the inequality.



"12" and "45" adding "33""1\not=2"\text{ and }"-4\not=-5"\text{ adding }\Rightarrow "-3\not=-3" :eek:

So, you're relying on something relating the terms, rather than just the general argument.
(edited 5 years ago)
Original post by ghostwalker
"12" and "45" adding "33""1\not=2"\text{ and }"-4\not=-5"\text{ adding }\Rightarrow "-3\not=-3" :eek:

So, you're relying on something relating the x1,x2x_1,x_2, rather than just the general argument.


I see your point. In which case, I think it's much easier to show the implication

F(x1)=F(x2)    x1=x2F(x_1) = F(x_2) \implies x_1 = x_2

to prove that F=gfF = g \circ f is injective.

@tom09099 I've edited my initial response regarding this.
(edited 5 years ago)

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