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Working with group theory I've found multiple times the idea of quotient group as $G/H = \{gH\ |\ g\in G, H < G\}$. Nevertheless, you can find similar things in vectorial spaces as $\mathbb{R}^2/L = \{P\vec{v} = \vec{v}^{\perp}\ |\ \vec{v} = \vec{v}^L + \vec{v}^{\perp},\ \vec{v}\in \mathbb{R}^2,\ \vec{v}^L \in L, \vec{v}^{\perp} \in L^{\perp}\}$. Here, $L$ is an straight line in $\mathbb{R}^2$

Is there a way to understand any quotient of this type without given the proper definition of the quotient (as I did above) nor equivalent relation? A general way to deduce what the quotient means by the meaning of the numerator and denominator?

Vicky
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    You can think of this in the framework of category theory, I suppose. –  Nov 20 '18 at 00:33
  • I don't get it. I've seen that $\mathbb{R}/\mathbb{Z} \simeq S^1$ and I don't see how to make the conection – Vicky Nov 20 '18 at 00:50
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    Please see my earlier answer on something similar giving some intuitive idea: https://math.stackexchange.com/questions/988522/intuitive-meaning-of-quotient-ring/988579#988579 – P Vanchinathan Nov 20 '18 at 00:57
  • Ok, then following your answer in that post, if I wanted to compute $\mathbb{Q}/\mathbb{Z}$, this would be ${0, 1/2}$ because I wouldn't be able to distinguish between two eleements in $\mathbb{Q}$ that differ by integers, true? And $\mathbb{R}/\mathbb{Z}$ would be $[0, 1]$, i.e., all real numbers between $0$ and one, taking them into account, true? – Vicky Nov 20 '18 at 01:10
  • Therefore, $\mathbb{R}/\mathbb{Z} \not\simeq S^1$ but $\mathbb{R}/(2\pi\mathbb{Z}) \simeq S^1$, isn't it? – Vicky Nov 20 '18 at 01:15

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