Math Proof: Surjective/Injctive

Question

$Q:$ Let $f : R → R$ be bijective. Define $g : R → R$ as $g(x) = |x|$. Then $g \circ f$ is neither surjective nor injective

Pf: Let $f: R → R$ be bijective. $g: R_1 → R_2$ as $g(x)=|x|$. Consider $x_1=2$ and $x_2=2$. $g(x_1)-g(2)=|2|=2=2=|-2|=g(-2)=g(x_2)$. Since $g(x_1)=g(x_2)$ then $g$ is not injective. Now consider $n=-2$. $-2$ does not exist in $R_2$ since $R_2\geq 0$ by the definition of absolute value and there does not exist and $x$ in $R_1$ s.t. $f(x)=-2$. As a result, $g$ is not surjective. So how do we prove that $g \circ f$ is neither surjective nor injective?

Answer 1

hint

for a real $x$, let

$$h(x)=g\circ f(x)=|f(x)|$$

thus

$$h(x_1)=h(x_2)\implies |f(x_1)|=|f(x_2)|$$

$$\implies f(x_1)=\pm f(x_2)$$

If $f$ is positive, then $g\circ f$ is bijective.

Answer 2

Firstly, it is clear that :→ is neither surjective nor injective.

Let $h$ be the inverse of $f$. Since, $f$ is a bijection, $h$ is also a bijection.

Now, assume that ∘ is an injective function. We also have that h is an injective function. Then, we can say (∘)∘$h$ is also an injective. But (∘)∘$h$ = ∘(∘$h$) = ∘$I$ = , which means g is injective. Contradiction! Thus the assumption that ∘ is injective is wrong. Similarly, ∘ can be proved as not surjective. Here, $I$ is the identity function.