$f:(0,1)\to \mathbb R^n$ continuous injective function s.t. $\lim_{x\to 0} f(x) , \lim_{x\to 1} f(x)$ doesn't exist , then $f(0,1)$ contractible?

Question

If $f:(0,1)\to \mathbb R^n$ be a continuous injective function such that $\lim_{x\to 0} f(x) , \lim_{x\to 1} f(x)$ doesn't exist , then is it true that $f(0,1)$ is contractible ?

Answer 1

ALERT: The following is only a try, which does not work.

Consider any homeomorphism $g:(0,1) \to \Bbb{R}$ for example $$t \mapsto \tan \left( \pi t - \frac{\pi}{2} \right)$$ Then consider the injective spiral $f:(0,1) \to \Bbb{C}$

$$f(t)= (1+e^{g(t)})e^{ig(t)}$$

Clearly, the limits $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$ do not exist. The image of $f$ is an infinite spiral wrapping around the unit circle (which never touches it), and it is not contractible.

EDIT: Sorry, this example does not work. I have in mind another one in $\Bbb R^3$, but it is too complicated to be described here. I hope someone else can give you a nice counterexample.

Answer 2

Suppose $f$ takes the interval $(0,1/4]$ to a decaying exponential (so its graph is the same as $e^{-x}$, but parametrized so that as $x \to 0$, it approaches its horizontal asymptote, while when $x=1/4$, $y=1$). Then $f$ takes the interval $[1/4, 1/2]$ to the quarter circle with radius 1 centered at the origin, between the points (0,1) and (-1,0). Finally, reflect the graph across the $x$-axis for the function on $[1/2, 1)$.

$f$ is injective. Is $f(0,1)$ contractible?