Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like a formal, but not very deep in the theory, answer to this question.

Maybe I am even wrong at the understanding of what a lemniscate may be, so here is another question:

Is the image of the function $f:[-2\pi,2\pi]\rightarrow \mathbb{R^2}$ given by:

$f(t)=(1+\cos(t),\sin(t))I_{(0,2\pi]}(t)+ (-1+\cos(t),\sin(t))I_{(-2\pi , 0)}(t)$

a lemniscate? (where $I_{X}$ is the indicator function of the set $X$)

share|cite|improve this question
Lemniscates are never manifolds because they cross over themselves and locally around the cross point it is not homeomorphic to $\mathbb{R}$ – Sean Ballentine Feb 19 '13 at 21:59
Yes, but why can't I get the homeomorphism? I understand the intuition, but as I said, I want something more formal. – Aloizio Macedo Feb 19 '13 at 22:04
up vote 12 down vote accepted

Suppose we try to chart near the cross point, so we have a homeomorphism $f:(-\epsilon,\epsilon) \to X \subset \mathbb{R}^2$ where $X$ is the small cross. Then if we remove the single point in the center of the cross in $X$ and the corresponding point in $(-\epsilon,\epsilon)$, we get 4 connected components in the image but only 2 in the pre-image. Since homeomorphisms must preserve the number of connected components, no such homeomorphism is possible.

share|cite|improve this answer
Formal answer and satisfactorily introductory. Upvoted and accepted, thanks! – Aloizio Macedo Feb 19 '13 at 22:12

As said in the comments, the lemniscate crosses over itself and thus cannot be homeomorphic around this point to $\mathbb{R}$. One way to see this is to think of what happens when the point is removed: you get 4 connected components in the lemniscate but only two in $\mathbb{R}$, so any map between the two cannot be an homeomorphism.

share|cite|improve this answer

Your set is the union of two touching circles. One can easily give an explizit homeomorphism with $\mathbb R^1$ (in fact by means of $f$) of a sufficiently small connected open neighbourhood of any point except the origon. But this is not possible (nor is it possible with any $\mathbb R^n$) for the origin:

Let $U$ be an neighbourhood of $(0,0)\in X\subseteq \mathbb R^2$, $V$ an open neighbouthood of $O\in\mathbb R^n$ and $h\colon U\to V$ a homeomorphism with $h(0,0)=O$. Wlog. $U=X\cap B_{\epsilon}(0,0)$ for some $\epsilon<1$. Then $U\setminus \{(0,0)\}$ has exactly four connected components, but the supposedly homeomorphic image $V\setminus \{O\}$ is connected (if $n>1$) or has two components (if $n=1$), contradiction.

share|cite|improve this answer
Why do we say "an neighborhood," instead "a neighborhood?" I've seen other people say that. I am starting to guess it’s on purpose. @Eitzen – Student Feb 19 '13 at 22:22
@George It really should be "a neighborhood". – Ragib Zaman Feb 19 '13 at 22:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.