A simple closed curve in $\mathbb{R}^{2}$ is a continuous map :
$$f : [0,1] \to \mathbb{R}^{2}$$
Where $f(0)=f(1)$ and $f$ restricts to $[0,1)$ is a injective map .
Is $f([0,1])$ homeomorphic to sphere $S^{1}$ ? Can I generalize this problem to higher dimension . For example : a convex polyhedron is homeomorphic to $S^{2}$
$S^1 = g[[0,1]]$, where $g: [0,1] \to \mathbb{R}^2: g(t) = \cos(2\pi t, \sin(2\pi t))$. $g$ also has the property that it is continuous and injective on $[0,1)$ and $g(0) = g(1)$. By compactness and Hausdorffness of all spaces involved, namely $Y:= f[[0,1]]$, $S^1$ and $[0,1]$, we know that the continuous $f$ and $g$ are quotient maps. Then $h: S^1 \to Y$ defined by $h((\cos(2\pi t), \sin(2 \pi t))) = f(t)$ and $h^{-1}: Y \to S^1$, with $h^{-1}(f(t)) = g(t)$ ($t \in [0,1])$ are well-defined, 1-1 and both are continuous as e.g. $h \circ g = f$ and $f$ is continuous and $g$ is quotient, etc. So $h$ is a homeomorphism.
Yes, it is. To prove it, one could use the standard construction of $S^1$ as the quotient space of $[0,1]$ where the end points are identified, and see that this is exactly what has happened to $f[0,1])$.
Yes. If $A$ is compact and $B$ is a Hausdorff space then a bijective and continuous map $f:A\to B$ has a continuous inverse.