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I've been trying to teach myself topology, and I'm having a bit of trouble grasping the abstract concepts of the field. One question that's been poking at my understanding regarding topological mappings is as follows:

Take an arbitrary mapping $ f $ (e.g. $ f(x) = x^2 $, the parabola in $ \mathbb{R}^2 $). Now, of course this mapping takes an argument from $ \mathbb{R} $, and returns an output also in $ \mathbb{R} $. We typically represent this by the ordered pair $ (x,f(x)) $. This ordered pair is an element of $ \mathbb{R} \times \mathbb{R} $ or just $ \mathbb{R}^2 $. Now should $ f $ be described as $$ f: \mathbb{R} \rightarrow \mathbb{R} \quad \mbox{or} \quad f: \mathbb{R} \rightarrow \mathbb{R}^2 ~? $$ My hypothesis is that it depends on if we want to analyze just the output of $ f $ or the ordered pair. I apologize in advanced for the trivial nature of the question.

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In general, when we write $f: A \to B$, we mean $f \subseteq A \times B$, with the condition that if $(x,a), (x,b) \in f$, then $a = b$. Since the second entry in the ordered pair is uniquely determined by $x$, we call this element $f(x)$. This condition is just saying that $f$ is well-defined.

We often think of $f$ as some machine that takes an element from the domain, and spits something of the codomain. But, having the caracterization of $f$ I said above, $f$ would be the same thing as its graph: $$\mathrm{gr}(f) = \{(x, f(x)) \in A \times B \ : \ x \in A \}$$

For practical purposes, we think of these thing separately, so I would go with $f: \Bbb R \to \Bbb R$, and $\mathrm{gr}(f) \subset \Bbb R^2$.

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I would say $f:\mathbb{R}\rightarrow \mathbb{R}$.

You could consider the related function $g:\mathbb{R}\rightarrow\mathbb{R}^2$ that is defined as $\{(x,(x,f(x))\mid x\in\mathbb{R}\}$

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The ordered pair is not $f$ it is the graph of $f$. So $f: \mathbb{R} \to \mathbb{R}$ and $G: \mathbb{R} \to \mathbb{R}^2$ where $G$ is the graphing function.

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