Context
Here's some context. I'm an undergrad physicist and this is my first post here, so please bear with me :)
Given a map, say $f: \mathbb{C} \to \mathbb{R}$ s.t. $f(z) = |z|$, ($f$ is clearly neither surjective nor injective), there is an induced equivalence relation on $\mathbb{C}$ : $z_1 \sim z_2$ if $f(z_1) = f(z_2)$, that is concentric circles of fixed radius. My understanding is that the inverse $f^{-1}: \mathbb{R} \to \mathbb{C}$ is a map the takes some radius and returns the subset of points forming the associated circle for example, $f^{-1}(a)$ is the set of points that form the circle of radius $a$ centered at the origin of the complex plane.
The quotient set $\mathbb{C}/\sim$ is then defined as the set of all equivalence classes, that is the set of all the circles centered at the origin. The quotient map $\tilde{f}:\mathbb{C}/\sim \to \{0\}\cup\mathbb{R^+}$ is the map the associates an equivalence class to its image $\tilde{f}([[x]]) = f(x)$, which in this example is taking a given given circle to its radius.
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Question
In all honesty, I don't understand the quotient set and map, so its hard to formulate a simple question. I wonder, since an element of an equivalence class is representative of all others, and since $\tilde{f}([[x]]) = f(x)$, why should there be any distinction between a map and its quotient? They seem to do the same thing, namely take a point (or a set of equivalent points) and return a positive real number. I can see the domain and the codomain of $f$ and $\tilde{f}$ are not the same, but this seems to be made arbitrarily so that $\tilde{f}$ is bijective on the range of $f$.
My notes don't mention any topology yet, so I'd like to avoid that for the time being. Thanks.
This
is a good question.
In a sense that you do in fact understand, there "is no distinction". That's because when you have a function $f: X \to Y$ there is a natural bijection between the partition constructed from inverse images of points in $Y$ to the range of $f$ (which may not be the whole codomain $Y$).
The "natural" in that assertion means you need make no arbitrary choices when defining it. What that bijection tells you is that you can use elements of the range to name the equivalence classes.
So if all you are thinking about is the structure of the function, you know everything when you know the partition. In any particular context there may well be more things you care about. For the function mapping a complex number to its length the fact that the length is a real number you can do something with may well matter. You can't easily do or interpret arithmetic on the circular equivalence classes.