Is the bijection of $3$ functions valid when you take them as a 3d vector? Let $F_1,F_2,F_3$ be three functions from $\mathbb R\to\mathbb R$. $F_1,F_2,F_3$ are bijective (we can say that they are strictly increasing on $\mathbb R$ by the bijection theorem they are bijective).
Let $g : \mathbb R^3\to\mathbb R^3 :=(F_1,F_2,F_3)$. Is $g$ bijective by composite?
 A: Yes
To prove that $g$ is bijective, we have to prove that it is injective and surjective.
Proof that $g$ is injective
Suppose that there is an element $y=(y_1, y_2, y_3)\in\mathbb{R}^3$ that is the $g$-image of more than one distinct element of $\mathbb{R}^3$. Then, there exist $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3)$ such that $g(u)=g(v)=y$. 
But this means that $F_1(u_1)=F_1(v_1)=y_1$. As $F_1$ is injective, we have $u_1=v_1$. The same reasoning can be applied to $F_2$ and $F_3$, yielding $(u_1,u_2,u_3)=(v_1,v_2,v_3)$ or $u=v$. This contradicts the premise that $y$ is the image of two distinct elements, so $g$ must be injective.
Proof that $g$ is surjective
Let $y=(y_1, y_2, y_3)\in\mathbb{R}^3$. As $F_1$ is surjective, we know that $y_1=F_1(u_1)$ for some $u_1\in\mathbb{R}$. Similarly, we know that $y_2=F_2(u_2)$ for some $u_2\in\mathbb{R}$, and that $y_3=F_3(u_3)$ for some $u_3\in\mathbb{R}$. If we let $u=(u_1, u_2, u_3)$, then we have $y=g(u)$. For any $y\in\mathbb{R}^3$, we can apply this procedure and find an $u\in\mathbb{R}^3$ such that $g(u)=y$. Therefore, $g$ is surjective.
