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Why do topological quotients "bend" lines?

http://mathonline.wikidot.com/topological-quotients-in-euclidean-space

I have no problem with the idea that one constructs a topology on the line from its subsets or that the equivalence relation $0 \sim 1$ "implies" that $0$ and $1$ must be connected. However, I don't understand what makes $(0,1)$ curved so that it forms a circle?

Gerry Myerson
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mavavilj
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    That's just a representation of the quotient space - there's no "bending" going on, and indeed there's no notion of curvature here at all: e.g. the circle is homeomorphic to the square. – Noah Schweber Sep 17 '18 at 01:14
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    Don't take that page too literally, there are several inaccuracies and severe errors! The definition of the final topology in the first example is just wrong. It should be as follows: Consider the canonical map $q:X\rightarrow X/\sim$, the final topology on $X/\sim$ is given by $\tau_{X/\sim}=\left{U\subset X/\sim \mid q^{-1}(U)\in \tau_X \right}$. Other than that, you don't construct any bending, rather you simply choose a way of visualizing these spaces. – Mathematician 42 Sep 17 '18 at 01:17
  • @NoahSchweber Unless the bending results from http://mathonline.wikidot.com/topological-quotients in which it's said that $X$ is a set and ${Y_i:i \in I }$ is a collection of topological spaces then the final topology induced by ${ f_i :i \in I }$ on $X$ is the finest topology $\tau$ which makes the maps $f_i : Y_i \rightarrow X$ continuous for all $i \in i$. -- Doesn't this read that the subintervals must connect continuously to each other, and since $0$ and $1$ are connected, then this "builds up" as arcs to form a circle. – mavavilj Sep 17 '18 at 01:17
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    @mavavilj No, there's still no bending - there's no notion of curvature at all. It's just a visualization: "$[0,1]/{0,1}$" is a bit abstract, but it's homeomorphic to the circle, which is familiar, so that's how we often represent it. There's really no bending going on here. – Noah Schweber Sep 17 '18 at 01:21
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    "Bending" or "curvature" is not a topological property and requires additional structure to be properly defined. – Mathematician 42 Sep 17 '18 at 01:21

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As has been mentioned in the comments, there is no "bending" going on here except in a very, very loose sense which is that no "straight" figure can be homeomorphic to the space in this particular example.

Rather, the relevant change here is from a space with boundary to one without. The choice of a circle to represent this is because it's an aesthetically pleasing shape (perfect symmetry) and very natural when considering how this topological space arises in other contexts, e.g. as the group of unit complex numbers, which have the natural structure of a circle. But from a purely topological point of view, a square, a rectangle, a Koch snowflake, or something considerably heegzbhier (e.g. a szheeeky zppikky thzing), are all equally valid representations - that's the whole point about topology: you throw all that detailed shape nitnoy stuff out the window and you are solely concerned with the relationships between points in terms of things like proximity and connectedness without specific distances or curvatures.

The_Sympathizer
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