0

$f \colon A \to \mathbb R$ be a function (where $A$ is some set) and define the function $g \colon A \to \mathbb R$ as $g(x) = 3 (f(x))^2 + 1.$ Prove if $g$ is injective then $f$ is injective

How do I prove the composite of the function is injective? I know for injection every $x$ value equals a y value $(x = y)$ for injection. Is this correct:

$3(f(y))^2 + 1= 3(f(x))^2 + 1 = f(y) = f(x)$

Thomas Andrews
  • 177,126
user3201346
  • 61
  • 4
  • 1
    Remember what injective means. Whenever $g(y) = g(x)$ then $x = y$. – Chris Cave Jul 30 '14 at 16:02
  • It isn't always true. Suppose $f(x) = x$ on $\mathbb{R}$, which is clearly injective (one-one). However $x^2 = (-x)^2$, so for your given $g(x)$ the composition is not injective. A general proposition is that the composition of injective functions is still injective, but your quadratic real polynomial $g(x)$ is not injective. – hardmath Jul 30 '14 at 16:04
  • This is no right $3(f(y))^2 + 1= 3(f(x))^2 + 1 = f(y) = f(x)$. Since $f(x)\not = g(x)$ – EQJ Jul 30 '14 at 16:21
  • 1
    You don't prove the composition of the functions is injective - that is a condition of the problem that you are supposed to assume is true. – Thomas Andrews Jul 30 '14 at 16:23

2 Answers2

1

You have that $g(x)$ is inyective. Then if you have $g(x)=g(y)$ then $x=y$. What does it mean on $f(x)$.

Suposse

$$f(x)=f(y) $$

then

$$g(x)=3f(x)^3+1=3f(y)^3+1=g(y)$$

which implies

$$x=y$$

Since $g$ is inyective.

in general if you have that $g=h\circ f$ is inyective then $f$ is inyective.

EQJ
  • 4,369
0

Since g is injective, we have $$\forall x,y \in A \,\,\, if \,\,\ g(x)=g(y) \to x=y \,\,\,\, (1)$$ Now assume $f(x)=f(y)$, we want to prove, in this case also $x$ equals to $y$. We have, $$f(x)=f(y) \to (f(x))^2=(f(y))^2 \to 3(f(x))^2=3(f(y))^2 \\ \to 3(f(x))^2+1=3(f(y))^2+1 \to g(x)=g(y)$$ Therefore, using equation (1), we conclude $x=y$, so $$f(x)=f(y) \to x=y$$ and therefore f is injective.

Mojtaba Golshani
  • 408