Why a vector function to $\mathbb{R}^n$ can be regarded as $n$ different functions

calculuselementary-set-theoryintuitionmultivariable-calculus

I am taking advanced calculus, and the professor said the following: we can regard $f: U \rightarrow \mathbb{R}^n$ as $n$ different functions, and write $f = (f_1,…,f_n)$."

I am not sure what the rigorous justification for representing the vector-valued function as $f = (f_1,…,f_n)$ (i.e. why we can write $f$ as $f = (f_1,…,f_n)$). How do we know that each component of the vector is a function of $p \in U$?

I thought of the following:

  1. Since $f: U \rightarrow \mathbb{R}^n$, then we know that $f(p)\in\mathbb{R}^n$, and thus we can write $f(p)$ as $f = (f_1,…,f_n)$, where $f_i$ are the components of $f(p)$.

  2. we can say that $f$ is comprised of $n$ different functions because when writing $f_i = \langle f, \mathbf{e}_i \rangle$, we have $n$ different functions from $U$ to $\mathbb{R}^n$, so (I hope so) this answers why we can think of $f$ as being made of $n$ functions.

  3. but why $f_i$ is function? how do we know that each component of $f$ is a function on its own? I think that we have already established this on the previous point – since $f_i = (f,\mathbf{e}_i)$, it is a composition of two functions ($(*,\mathbf{e}_i): U \rightarrow \mathbb{R}$ composed over $f$), and thus is a function (or alternatively, say that $f_i = \pi_i \circ f$ which is again a function since it is the composition of the projection function over $f$).

But I feel like I lack the "intuitive" understanding of why this is true… the only intuitive (and contorted) explanation I came close to is the following: Since $f$ is a vector-valued function, we know that $f(p) \in \mathbb{R}^n$, and since it is a vector, we know that the vector and its components are equivalent. Since we know that $p$ determines $f(p)$ and that it is equivalent to its components because it is a vector, we can say that each of its components is also a function of $p$.

I would be grateful if you could make these points clearer (and maybe provide a formal definition, as I think that the cause for my misunderstanding is the lack of a formal definition. I looked for formal definitions in the literature (Ruding for example, but since it is very basic I wasn't able to find any rigid and rigorous definitions there).

Best Answer

To justify this precisely, we first need to define what $(f_1,\dots,f_n)$ means. Usually, mathematicians don't go into this level of detail, or else assume that the reader will be able to fill in the gaps themselves, which perhaps explains the root of your confusion.

If $U$ is any set (usually, but not necessarily, a subset of $\mathbb R^m$ for some $m\ge0$), and $f_1,\dots,f_n$ are functions from $U$ to $\mathbb R$, then we define $(f_1,\dots,f_n)$ as the map $U\to\mathbb R^n$ given by $$ (f_1,\dots,f_n)(x)=(f_1(x),\dots,f(x)) \, , $$ for all $x\in U$. Conversely, if $f$ is a function $U\to\mathbb R^n$, where $n\ge0$, then we write $f_i$ as a shorthand for $\pi^i\circ f$, where $\pi^i$ denotes the projection of $\mathbb R^n$ onto the $i$-th coordinate, given by $\pi^i(x_1,\dots,x_n)=x_i$. (Technically, this is a different function for each value of $n$, so if we were being very fussy, we should write something like $\pi_n^i:\mathbb R^n\to\mathbb R$ instead.)

Now we can state and prove the following theorem:

Theorem. With the notation as above, for all sets $U$ and $n\in\mathbb N$ and functions $f:U\to\mathbb R^n$, we have $f=(f_1,\dots,f_n)$.

Proof. Let $x\in U$, and write $f(x)=(y_1,\dots,y_n)$. Then, \begin{align} (f_1,\dots,f_n)(x)&:=(\pi_1\circ f,\dots,\pi_n\circ f)(x) \\ &= ((\pi_1\circ f)(x),\dots,(\pi_n\circ f)(x)) \\ &= (y_1,\dots,y_n) \\ &= f(x) \, , \end{align} from which the result follows.

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