A question about Folland's definition of parametrization

Question

Here is Folland's definition of parametrization in his book Real analysis: modern techniques and their applications Chapter 11.2 Hausdorff measure.

We now consider lower-dimensional sets in $\mathbb{R}^n$. If $1 \le k \le n$, a k-dimensional $C^1$ submanifold of $\mathbb{R}^n$ is a set $M\subset\mathbb{R}^n$ with the following property:

For each $x \in M$ there exist a neighborhood $U$ of $x$ in $\mathbb{R}^n$, an open set $V\subset\mathbb{R}^k$, and an injective map $f:V\to U$ of class $C^1$ such that $f(V) = M\cap U$ and the differential $D_xf$ - i.e., the linear map from $\mathbb{R}^k$ to $\mathbb{R}^n$ whose matrix is $\left[\left(\frac{\partial f_i}{\partial x_j}\right)(x)\right]$ - is injective for each $x\in V$. Such an $f$ is called a parametrization of $M \cap U$.

Compare this definition with the definition of smooth manifold, which requires the coordinate map is a local homeomorphism.

I want to show in Folland's definition, the local inverse $f^{-1}:M\cap U\to V$ is continuous with respect to the subspace topology of $M\subset\mathbb{R}^n$.

But in Do Carmo's book Differential geometry of curves and surfaces, in the definition of regular surface, local homeomorphism is required.

Thanks in advance if you can give me any hint (or counterexample)!

Thanks to @Kenny Wong's answer (figure 8 space) below. We can conclude that, "figure 8 space" is a $1$-dimensional $C^1$ manifold in Folland's definition, but in the mean time, it is not a topological manifold if it is equipped with subspace topology.

Answer 1

Pick $p \in V$, and pick an open ball $B(p, \delta)$ such that its closure $\overline{B(p, \delta)}$ lies inside $V$.

Consider the restriction of $f$ to the closed ball $\overline{B(p, \delta)}$. This is a continuous injective map from a compact space to a Hausdorff space. Therefore, this map is a homeomorphism onto its image.

The restriction of this map to the open ball $B(p, \delta)$ is then also a homeomorphism onto its image.

So at the very least, the following statement is true:

For every $p \in V$, there exists an open neighbourhood $V'$ of $p$ with $V' \subset V$ such that $f|_{V'} : V' \to f(V')$ is a homeomorphism.

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However, taking your definition as written, it is not necessarily true that $f: V \to f(V)$ is itself a homeomorphism.

For example, suppose that $V = (-\pi, \pi)$ and suppose that $U = \mathbb R^2$. Let $f : V \to U$ be the "figure-of-eight" map, $$ f(t) = (\sin 2t, \sin t).$$ Take $M$ to be its image, $f(V)$. By your definition, $M$ is a $1$-dimensional submanifold of $\mathbb R^2$.

Please sketch this to get a feel for what this looks like, and observe that $f$ is not a homeomorphism onto its image, since the inverse $f^{-1}|_{f(V)}$ is discontinuous at $(0, 0)$.

[That said, the restrictions $f|_{(-\delta, \delta)}$, $f|_{(\pi - \delta, \pi)}$ and $f|_{(-\pi, -\pi + \delta)}$ are all homeomorphisms onto their respective images, so there is no contradiction with what we said earlier.]