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?
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.