Show that a set fails to be a manifold

Question

Spivak defines manifolds in this way:

A subset $M$ of $\mathbf R^n$ is called a $k$-dimensional manifold (in $\mathbf R^n$) if for every point $x\in M$ the following condition is satisfied:

$(M)$ There is an open set $U$ containing $x$, an open set $V\subset \mathbf R^n$, and a diffeomorphism $h:U\to V$ such that $$h(U\cap M)=V\cap(\mathbf R^k\times\{0\})=\{y\in V: y^{k+1}=\dots=y^n=0 \}.$$

And gives this theorem:

5-2 $\,\,\,$ Theorem. $\textit{A subset}$ $M$ $\textit{of}$ $\mathbf{R}^n$ $\textit{is a}$ $k\textit{-dimensional manifold}$ $\textit{if and}$ $\textit{only if}$ $\textit{for each point}$ $x\in M$ $\textit{the following}$ $\textit{“coordinate condition”}$ $\textit{is satisfied:}$

$\quad (C)$ $\textit{There is}$ $\textit{an open set}$ $U$ $\textit{containing}$ $x\textit{, an}$ $\textit{open set}$ $W\subset\mathbf{R}^k\textit{,}$ $\textit{and a}$ $1\textit{-}1$ $\textit{differentiable}$ $\textit{function}$ $f:W\to\mathbf{R}^n$ $\textit{such that}$ $$\begin{align}(1) \ & f(W)=M\cap U,\\ (2)\ & f'(y)\textit{ has rank }k\textit{ for each }y\in W,\\ (3) \ & f^{-1}:f(W)\to W\textit{ is continuous}.\end{align}$$

He then asks to make up a counterexample to Theorem 5-2 if condition $(3)$ is omitted and as a hint says "wrap an open interval into a figure six". But I don't have any idea on how to do that, nor do I have other ideas for counterexamples.

The only idea that I had is to translate Spivak's condition $(C)$ into more comprehensible (i.e., more comprehensible/familiar for/to me) language (see my other question) and then figure out what condition $(3)$ means in terms of the setting which I'm familiar with. But probably it's easier to do this problem directly, I don't know.

Answer 1

See this question. The interval $(-\pi,\pi)$ is "wrapped" to form a figure eight via the map $(-\pi,\pi) \to \mathbb{R}^2$, $t \mapsto (\sin 2t, \sin t)$. This map is one-to-one and differentiable.

However, the inverse map from the image to $(-\pi,\pi)$ fails to be continuous for open neighborhoods of the "crossover point." This is easiest to think about in terms of limits. There are sequences in the image converging toward the crossover point, but the termwise preimage of the sequence coverges to $\pi$ or $-\pi$, not the preimage of the crossover point, which is zero.

EDIT: Answer to question in comment. Let $a_n$ be a sequence in $(-\pi,\pi)$ converging to $\pi$, for example $a_n = \pi - \frac{1}{n}$. Then $f(a_n)$ is a sequence in the figure-eight converging to the origin. Note that $f$ is injective, so it is bijective when we restrict the range to be the image of $f$. Also note that $f(0)$ is the origin. If $f^{-1}$ is continuous, then $$ \lim_{n \to \infty} f^{-1}(f(a_n) ) = f^{-1} \big(\lim_{n \to \infty} f(a_n)\big) $$ The LHS is $\lim a_n = \pi$, but the RHS is $f^{-1}(0,0) = 0$, so $f^{-1}$ can't be continuous.

Answer 2

Consider$$\begin{array}{rccc}f\colon&(-1,1)&\longrightarrow&\mathbb{R}^2\\&t&\mapsto&\begin{cases}(t,0)&\text{ if }t<0\\\left(\cos\left(-\frac\pi2+2\pi t\right),1+\sin\left(-\frac\pi2+2\pi t\right)\right)&\text{ otherwise.}\end{cases}\end{array}$$Then $f\bigl((-1,1)\bigr)$ is the union of the segment joining $(-1,0)$ to $(0,0)$ together with the circle centered at $(0,1)$ with radius $1$. It is not a manifold because of the point $(0,0)$. If you remoe it, then, yes, it is a manifold.