I am trying to solve a homework problem where a quotient space is defined in a way that I do not understand, so I will ask a more simplified question.

I understand that $$ [0,1]/0 \sim 1, $$

means that you sew the points 0 and 1 together, getting a circle. The equivalence classes are all singletons except for the one that contains 0 and 1. There is also the common example where a ball is converted into a sphere by sewing the boundary of the ball together.

But what does the quotient space defined by

$$ [-1, 1]/[0,1/2]$$

look like? The notation is telling me that all points $[0,1/2]$ are identified to a single point, but I still don't understand what this operation means exactly.

I drew a sketch of what I thought happened when $[0,1/2]$ was identified to a single point:

enter image description here

I would really like to understand what this space looks like, and what "identified to a single point" means for this problem.

  • $\begingroup$ How is $\sim$ defined? $\endgroup$ – Git Gud Mar 17 '13 at 17:25
  • $\begingroup$ @GitGud It identifies the two points - OP describes it in the next sentence. $\endgroup$ – us2012 Mar 17 '13 at 17:27

It’s much simpler than your picture: it just contracts the interval $\left[0,\frac12\right]$ to a single point. The resulting space is homeomorphic to a closed interval of $\Bbb R$. Let $p$ be the point of the quotient corresponding to the interval $\left[0,\frac12\right]$. Perhaps the most obvious homeomorphism is

$$h:[-1,1]/[0,1/2]\to[-1,1/2]:x\mapsto\begin{cases} x,&\text{if }-1\le x<0\\\\ 0,&\text{if }x=p\\\\ x-\frac12,&\text{if }\frac12<x\le 1\;. \end{cases}$$

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  • $\begingroup$ I guess I am getting too hung up on visualization. $\endgroup$ – Mr. Fegur Mar 17 '13 at 17:57
  • 1
    $\begingroup$ @Mr.Fegur: Just think of this one as making the interval $\left[0,\frac12\right]$ shrink to a single point, pulling the intervals $[-1,0)$ and $\left(\frac12,1\right]$ together to meet at that point. But yes, sometimes it’s pretty nearly impossible actually to visualize a quotient. $\endgroup$ – Brian M. Scott Mar 17 '13 at 18:03

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