Embedding and not a homeomorphism In a different thread, I came across the following function $f: (0, 1) \mapsto \mathbb{R}^2 $ (proposed by Christian Blatter):
$$ f(t) = \begin{cases} (6t-1, 0) &\text{ if } 0 < t \le \frac{1}{3}\\
                        (2-3t, 3t-1) &\text{ if } \frac{1}{3} < t \le \frac{2}{3}\\
                        (0, 3-3t) &\text{ if } \frac{2}{3} < t < 1.
          \end{cases}
$$
The plot of the function is presented below.

I can see that, as a function from $(0, 1) $ to $\mathbb{R}^2, $ $f $ is not a homeomorphism since $f^{-1} $ fails to be continuous. However, I was wondering if the function is still an embedding. It seems that when we consider the function $f: (0, 1) \mapsto f((0,1)) $ and equip $f((0,1)) $ with the relative topology inherited from $\mathbb{R}^2 $ things are a little bit more under control.  
Thank you,
Maurice
 A: You seem somewhat confused. We have $f: (0,1) \rightarrow \mathbb{R}^2$, which is continuous and one-to-one (because $0$ and $1$ are not part of the domain). It is not surjective (not all points of the plane are in its image), so for that reason alone $f$ is not a homeomorphism.
It then still could be an embedding. We consider the same function $f$ as a map between $(0,1)$ and $f[(0,1)]$, which is by definition surjective (it should really have a different name then, to avoid confusion, but it's common to use the same letter for both). $f$ is an embedding whenever this restricted map (restricted in the image) is a homeomorphism. (This is the definition of an embedding, essentially). This would make $(0,1)$ and $f[(0,1)]$ homeomorphic, and the latter would be a so-called embedded "copy" (topologically) of the former.
However, in that case the inverse function $f^{-1}: f[(0,1)] \rightarrow (0,1)$ would also have to be continuous. But consider $x_n = (0,\frac{1}{n}) = f(1-\frac{1}{3n})$ which converges in the plane (and $f[(0,1)]$) to $(0,0) = f(\frac{1}{6})$. So $x_n \rightarrow (0,0)$ in $f[(0,1)]$, but $f^{-1}(x_n) = 1 - \frac{1}{3n}$ does not converge to $f^{-1}(0,0) = \frac{1}{6}$ in $(0,1)$. So $f^{-1}$ does not preserve convergent sequences and cannot be continuous.
Intuitively, it's clear that the image of $f$ does not resemble an interval, due to its "three-meeting" point $(0,0)$. E.g. in $(0,1)$, removing any point leaves the remainder disconnected, while we can remove $(1,0)$ from $f[(0,1)]$ and have a connected remainder. Arguments like this can also be used to show that $(0,1)$ and its image under $f$ are not homeomorphic so that $f$ is not an embedding.
