Why isn't a lemniscate a manifold? 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 $X$)
 A: 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.
A: 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.
A: 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.
