Heuristic for quotient topology: Is it really a gluing? In every book they say: gluing points together in a topological space is done via the quotient topology. Then they provide some examples, but they don't try to explain WHY it is true for any kind of example we can think of:

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*Considering $[0,1]/\sim$ with the quotient topology identifying $0$ with $1$ leads to a homeomorphism to the circle. That's also what our imagination says!

*Another examples is identifying the left and right site of a rectangle in the same direction, which leads to $[0,1]^2/\sim$ with the quotient topology, which is homeomorphic to a cylinder. That's also what our imagination says!

*Again, another example is considering the closed disk identifying the boundary $\mathbb{D}/\sim$ and using the quotient topology to get a homeomorphism to the sphere $\mathbb{S}^2$. That's also what our imagination says!

My question:
If we consider an object $X\subset \mathbb{R}^3$ and want to glue points together, our imagination in our head(!) leads to an object $Y\subset \mathbb{R}^3$ (just glue the points in your head together). But the books always say: Consider $X/\sim$ and take the quotient topology on it.
What is a heuristic argument, that $X/\sim$ is homeomorphic to $Y$? I want a "why it should be true quotient topology leads really to gluing:..." and not just "your three examples provide evidence, that is good enough and should be true for all other cases...".
EDIT: Everyone of you stuck at the point, that $Y$ may not be in $\mathbb{R}^3$. That is not the point of my question. Just restrict your attention to examples, where $Y$ is in $\mathbb{R}^3$.
I don't want to take the definition for granted. I want to understand them and convince myself that's how we should do it. If you do research, you also have to find the right definition to capture the behavior you want. That may be the most difficult part in mathematical research.
MAIN QUESTION:
If you never heard of the quotient topology (say we developed the theory of topological spaces one week ago), but you want to formalize the concept of gluing. You have a topological space $(X,\tau)$. Now you have an equivalence relation $\sim$ on $X$ and want to get a topological space $(X/\sim, ?)$.
How would you come up with the right topology on the quotient set $X/\sim$ to catch the behavior of gluing in a mathematical precise manner?
 A: Your question is too vague to give a precise answer to. After all, I don't understand what your Y is or why you think it was assembled by gluing or why it still sits in $\Bbb R^3$. Until you can tell me what gluing is I can't tell you why that's what quotients are.
Still, here is the result that unifies all of your three examples. I leave the proof as an exercise to you.

If $X$ is a compact space, and $Y$ a Hausdorff space, and $f: X \to Y$ is a continuous surjection, then $f$ descends to a homeomorphism $X/\sim \to Y$, where $x \sim y$ if $f(x) = f(y)$.

Whatever your space Y is, when you tell me that you are imagining gluing points of X together, you are presumably telling me a map from X to Y. That's what the above is. Then the result says that Y is precisely the quotient space obtained by identifying the points in X that f does.
In your first example, f traces out the circle counterclockwise. The second example is similar. In your third example, f stretches a rubber disc over the 2-sphere with the entire boundary of that disc getting mapped to the bottom point.
A: You can’t always glue things together and remain in $\mathbb R^3.$ For example, if you have a Möbius strip, identify points on the boundary can give the Klein bottle, which cannot be embedded in $\mathbb R^3.$ Same for the real projective plane, which can be gotten by gluing a disk to the Möbius strip.
Both spaces require four dimensions to embed them.
So your intuition is itself broken.
“Gluing” implies a mechanical process, and it sometimes matches that intuition. But not always. There is no way to actually glue a disk to a Möbius strip, if you are dealing with real-world models. (To me, your third example doesn’t match my intuition of a “gluing,” but your mileage may vary.)
Showing that in general, quotients/gluings match your intuition is kind of impossible. You can only convince yourself with individual cases over time.
In reality, spaces like the torus, the sphere, the Klein bottle and the Möbius strip exist as entities independent of what dimension you visualize them in.
Indeed, topologists think of all four of these spaces as fundamentally $2$-dimensional, because locally they all are topologically the same as the plane (or the half-plane, on the boundary of the Möbius strip.) The properties of these spaces are independent of where they live.
The game “Asteroids” takes place on a torus, because the bottom of the screen and top of the scree are the same, and left and right are the same. It doesn’t help you navigate the game to think of it as a three-dimensional space. It is fundamentally two-dimensional.
A: For intuition on how open sets encode "nearness", consider the slightly more concrete setting of metric spaces, where openness is defined in terms of distance. In a metric space $M$, the following are equivalent:

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*$U$ is open.

*$U$ is a union of open balls (sets of the form $B(x;r)=\{y\in M:d(x,y)<r\}$, with $r>0$).

*For every $x\in U$, there is some $r>0$ such that $B(x;r)\subseteq U$.

*$U$ includes a neighborhood of each of its points.

*Every element of $U$ is an interior point of $U$.

Even though not every topological space is metrizable and there are spaces that violate our intuition, the metric space "picture" of open sets still motivates a lot of the definitions.
To define the quotient topology, we have a topology on $X$ and we want to specify the open sets of $Y=X/{\sim}$ so as to encode the "nearness of points" that results from gluing according to $\sim$. So for each set $S\subseteq Y$, we need to decide whether $S$ is open. This reduces to deciding what it means to be an interior point of $S$.
Suppose $y\in S$ is a glued point, i.e. $q^{-1}(y)=\{a,b\}$, where $q:X\to Y$ maps each point to its $\sim$-equivalence class. Since we've glued $a$ and $b$ together to make $y$, the new "neighbors" of $y$ are the points in $Y$ that come from "neighbors" of either $a$ or $b$ in $X$. For $y$ to be an interior point of $S$, $S$ must "include all of $y$'s neighbors", so we need $q(x)\in S$ whenever $x$ is "near enough" to $a$ or $b$. In other words, $q^{-1}(S)$ must include a neighborhood of $a$ and a neighborhood of $b$. In other words, $a$ and $b$ must be interior points of $q^{-1}(S)$.
Generalizing this to arbitrary $y$, we get that $y$ is an interior point of $S$ iff each element of $q^{-1}(y)$ is an interior point of $q^{-1}(S)$. Then we have the following chain of equivalent statements:

*

*$S$ is open (in $Y)$.

*Every element of $S$ is an interior point of $S$.

*For every $y\in S$, every element of $q^{-1}(y)$ is an interior point of $q^{-1}(S)$.

*Every element of $q^{-1}(S)$ is an interior point of $q^{-1}(S)$.

*$q^{-1}(S)$ is open (in $X$).

so we've arrived at the definition of the quotient space topology.
