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I see this pretty much everywhere, but I'm still not really sure what it means. I've heard the explanation that it "maps the third dimension onto the second dimension," but even that doesn't make sense to me. Could someone elaborate?
An example: Suppose that $F : \mathbb{R}^2\rightarrow\mathbb{R}$ and $f : \mathbb{R}\rightarrow\mathbb{R}$ are differentiable functions...
Thanks!

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    $\begingroup$ I have no clue what you're talking about or what your question is $\endgroup$
    – Ant
    Commented Dec 15, 2015 at 18:11
  • $\begingroup$ It's a function from $\mathbb{R}^3$ to $\mathbb{R}^2$. Are you unsure about what $\mathbb{R}^n$ means? $\endgroup$
    – anomaly
    Commented Dec 15, 2015 at 18:12
  • $\begingroup$ it means you take three inputs and produce only two $\endgroup$ Commented Dec 15, 2015 at 18:12
  • $\begingroup$ Thanks for the clarification @Ant... I think I get it now, good answers/comments! Thanks all. $\endgroup$ Commented Dec 15, 2015 at 18:17

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When someone writes $$f : X \rightarrow Y $$ They are using notation to explain something about the function $f$. In particular this spells out what the domain and codomain spaces are. For $$f : \Bbb{R}^3 \rightarrow \Bbb{R}^2$$ it means that the function takes input from $\Bbb{R}^3$ and has output values in $\Bbb{R}^2$. For example $$f(x,y,z) = (x^2 - y^2, 3z - 2x + y)$$

This notation is used to clarify, right away, what to expect about the input and outputs of a function.

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An application $f$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ is a function that for all vectors of $\mathbb{R}^3$ gives one (and only one) vector of $\mathbb{R}^2$.

There are a lot of examples:
$f=0$. This map sends every $(x,y,z)$ to $(0,0)$
$f(x,y,z) = (x,y)$. This map ignores the third coordinate.
$f(x,y,z) = (x+y+z, 2y)$. This map... does stuff.
and so on.

If you're in the context of vector spaces, $f$ is required to be linear: this means that for all $v,w \in \mathbb{R}^3,$ $a,b \in \mathbb{R}$ $$f(av+bw) = af(v)+bf(w)$$ and, if this holds, it can be represented by a $3\times 2$ matrix.

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This means you take three inputs: $(x,y,z)$ and only produce two outputs.

$$f(x,y,z) = (g(x,y,z),h(x,y,z))$$ as in

$$f(x,y,z) = (x^2zy,xy^2z)$$

you begin with an element of $\mathbb{R}^3$ and output an element of $\mathbb{R}^2$.

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  • $\begingroup$ Two ... outputs, to finish the sentence. Of course an alternative interpretation is that a 3-tuple counts as a single input and a 2-tuple counts as a single output. $\endgroup$
    – anon
    Commented Dec 15, 2015 at 18:18
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    $\begingroup$ @anon Naturally, but this may be easier for the OP to think about. $\endgroup$ Commented Dec 15, 2015 at 18:19
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A slightly more formal explanation: a member of $\mathbb{R}^3$ can be represented as a vector with three (real) components $\langle x,y,z\rangle$ and a member of $\mathbb{R}^2$ similarly is a vector with two components $\langle v,w\rangle$. A function $f():\mathbb{R}^3\mapsto\mathbb{R}^2$ is then a mapping from the first of these spaces to the second; for clarity's sake we often write $f(\langle x,y,z\rangle)$ as $f(x,y,z)$, and there's a natural isomorphism between the two, but it can often be useful to think of $f()$ as a function with one argument. For instance, the length function $f():\mathbb{R}^3\mapsto\mathbb{R}$ can be written as $f(x,y,z) = \sqrt{x^2+y^2+z^2}$ but it's often both easier and more insightful to write $f(\overrightarrow{v})=|\overrightarrow{v}|$, where the argument of the function is explicitly the vector $\overrightarrow{v}$.

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