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I am working on this question:

Considering the elipse: $E=x^2/4+y^2=1$ and the equivalence relation $(x_1,y_1)R(x_2,y_2)$ if and only if $x_1^2+y_1^2=x_2^2+y_2^2$.

Is $(E/R,\epsilon_1/R)$ homeomorphic to $([1,2],\epsilon_1)$? $\epsilon_1$ is the usual topology over $ \mathbb R $

The equivalence relation tells me each equivalence class is formed of 4 points with the coordinates $(\pm x,\pm y)$ over the ellipse, so it's like the ellipse gets transformed to the arc of ellipse in the first quadrant. I have already shown that the space is connected and compact.

1) First of all, is there an error is this question? Since the ellipse leaves in $\mathbb R^2$, shouldn't it be $(E/R,\epsilon_2/R)$? or is it ok because it is a line an when restricting to it, it is in some sense a one-dimensional thing (you can only go forward or backward along it) ?

2) Intuitively it looks like it is fact an homeomorphism, but I am not sure how to prove it. Any idea?

emacs drives me nuts
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some_math_guy
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1 Answers1

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1) This depends on what the author meant by $\epsilon_1$. Either way the context is clear: the topology is supposed to be inherited from $\mathbb{R}^2$. So whether the index is $2$ or $1$, is it typo or not, is really irrelevant.

2) Assuming you've correctly concluded that only tuples of the form $(\pm x,\pm y)$ are related (which seems correct) we have

$$q:[0,2]\to E/R$$ $$q(r)= \bigg[r, \sqrt{(1-r^2/4)}\bigg]_R$$

which is a homeomorphism. Continuity is simple because the function factorizes through $E$. You can check that under the relationship this is a bijection and thus a homeomorphism (because $[0,2]$ is compact).

freakish
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  • Continuity follows from multivariable calculus, but what do you meant by" the function factorizes through E"?. and what does knowing [0,2] is compact used for? – some_math_guy Feb 07 '20 at 13:21
  • @juancarlosvegaoliver I meant that $q=\pi\circ f$ where $\pi:E\to E/R$ is the quotient map and $f:[0,2]\to E$ is given by $f(r)=(r, \sqrt{1-r^2/4})$. Note different brackets. Without compactness we only get that $q$ is a continuous bijection. – freakish Feb 07 '20 at 13:51
  • What's the general statement of this theorem? – some_math_guy Feb 07 '20 at 13:59
  • @juancarlosvegaoliver A continuous bijection from a compact Hausdorff space to a Hausdorff space is a homeomorphism: https://math.stackexchange.com/questions/3042668/continuous-bijection-between-compact-and-hausdorff-spaces-is-a-homeomorphism – freakish Feb 07 '20 at 13:59
  • While the closed interval is a compact subset, the whole space $ \mathbb R $ is not compact, the theorem requires a compact space – some_math_guy Feb 07 '20 at 14:03
  • @juancarlosvegaoliver the domain of $q$ is $[0,2]$, not $\mathbb{R}$. I don't care about $\mathbb{R}$ here. I'm not even sure why you started to talk about $\mathbb{R}$... – freakish Feb 07 '20 at 14:04