how to prove a parametric relation to be a function For example lets suppose that I have given the functions $f:\mathbb{R}\longrightarrow \mathbb{R}$ and $g:\mathbb{R}\longrightarrow \mathbb{R}$. If my relation is $R=\{(x,(y,z))\in \mathbb{R}\times \mathbb{R}^{2}: y=f(x) \wedge z=g(x)\}$ How to prove formally (from a set theoretic stand point) that $R$ is a function. I have a try but I'm not convince: 
Let suppose to have $(x,(y,z))\in R$ and also $(x,(y',z'))\in R$. Then $y=f(x), z=g(x)$ and also $y'=f(x), z'=g(x)$ by definition. Then $y=y'$ and also $z=z'$. Therefore $(y,z)=(y',z')$. 
In a more general case if I have the functions $f_1, f_2,...,f_n:\mathbb{R}^m\longrightarrow \mathbb{R}$ and I define the function $f:\mathbb{R}^{m}\longrightarrow\mathbb{R}^{n}$ such that $f(x_{1},x_{2},...,x_{m})=(f_{1}(y),f_{2}(y),...,f_{n}(y))$ with $y=(x_1,x_2,...,x_m)$, how to justify that it is indeed a function? If my try is fine I suppose this can be done by induction. Any comment will be appreciated. 

Let's prove a vastly more general statement.

Let $I$ be an index set, and for every $i\in I$ let $f_i$ be a function. Then the relation $F$ defined on $X=\bigcap_{i\in I}\operatorname{dom}(f_i)$ by $F=\{\langle x,\langle f_i(x)\mid i\in I\rangle\rangle\mid x\in X\}$ is a function.

Proof. Let $x\in X$, and suppose that $\langle x,\langle y_i\mid i\in I\rangle\rangle,\langle x,\langle z_i\mid i\in I\rangle\rangle\in F$, then for every $i\in I$ we have $y_i=f_i(x)=z_i$, therefore the sequences are equal and $F$ is a function. $\square$

Now you care about the case where $\operatorname{dom}(f_i)$ are all equal, so the intersection creating $X$ is trivial.
