I gave myself a claim to prove:
If $f: U \rightarrow \mathbb{R}^n$, then we can write $f$ as $f=(f_1,…,f_n)$.
I tried to prove it this way:
Assume $f: U \rightarrow \mathbb{R}^n$. Then we know that $f(p) \in \mathbb{R}^n$. So we can write $f(p) = (x_1,…,x_n)$. We can proceed in two ways:
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Since $f(p) = (x_1,…,x_n)$, each $x_i$ depends on $p$, and so it is a function of $p$, and thus each $x_i$ is a function $x_i: U \rightarrow \mathbb{R}$. This line of "proof" seems not formal enough and doesn't show rigorously that $x_i$ is a function from $U$ to $\mathbb{R}$.
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Since $f(p) \in \mathbb{R}^n$, we can define $f_i \:= f \cdot \mathbf{e}_i$ (which is a function from $U$ to $\mathbb{R}$), and then $f = (f,\mathbf{e}_i)\mathbf{e}_i = (f_1,…,f_n)$ (I used Einstein summation convention), where $\mathbf{e}_i$ is the standard basis of $\mathbb{R}^n$.
I know it is a very basic proof, but I am not confident in my proof-writing skills yet, so I would be grateful if you could provide your input and advice on how to make it more rigorous and precise.
Best Answer
I think that what you are asking should be restated as another question that might make things clearer: "Why can $f: U \rightarrow \mathbb{R}^n$ be regarded as n different functions? (why can we say that f is comprised of n different functions?)"
With this formulation of the question, it might be less vague and easier to answer your question: Since $f: U \rightarrow \mathbb{R}^n$, we know that it is a vector in $\mathbb{R}^n$ (and as such, we know it has $n$ components or coordinates), so we can decompose $f$ to $n$ distinct functions: $f_i: U \rightarrow \mathbb{R}^n; f_i = (f,\mathbf{e}_i)$, or as suggested by Mr. Giorgos Giapitzakis, you can use the projection mapping $\pi_i:\mathbb{R}^n \rightarrow \mathbb{R}; \pi_i(\vec{x})=x_i$. Those are $n$ different functions of which f is comprised, and they are exactly the coordinate function of $f$ (if you want to go further in detail, you can ask why each component/coordinate of the image vector is a function on its own. You can go about it intuitively and more formally. Formally, we already showed that each of the coordinates is a function, because we showed that $f_i = (f,\mathbf{e}_i)$, which as said before is a function from $U$ to $\mathbb{R}$. Intuitively, we can say that since each $p$ determines a vector $f(p)\in \mathbb{R}^n$, and because each vector determines its coordinates (and vice-versa), and thus each coordinate of the vector depends on p on its own. Why does it make sense that each of the n coordinates depends on $p$ / is a function of $p$? It makes sense because $f$ gives us for every $p$ a vector, meaning it returns $n$ values. Now it means that each of the $n$ components gets a value for each $p\in U$, and hence each coordinate is a function of $p$).