Do simple paths have unique endpoints? Let $Y \subset \mathbb{R}^n$ be a simple path.
That is there exists a continuous injective function $f : [0, 1] \rightarrow \mathbb{R}^n$ such that the image of $f$ is $Y$.
Note, by this definition $f(0)$ cannot equal $f(1)$.
Now assume there exists another continuous injective function $g : [0, 1] \rightarrow \mathbb{R}^n$ whose image is also $Y$.
Can $\{f(0), f(1)\} \neq \{g(0), g(1)\}$?
 A: Let's take ourselves through Greg's suggestion. If $g : [0, 1] \to Y \subseteq \Bbb{R}$ is injective, then it is bijective due to restricting the range. This implies $g^{-1}$ exists. We wish to show it is continuous.
Suppose $(x_n) \in Y$ is a sequence converging to $x \in Y$. Let $t_n = g^{-1}(x_n)$ for all $n$, and $t = g^{-1}(x)$. We wish to show that $t_n \to t$.
Suppose, for the sake of contradiction, that $t_n \not\to t$. Then, for some $\varepsilon > 0$, there must exist a subsequence $t_{n_k}$ such that $|t_{n_k} - t| \ge \varepsilon$. By taking further subsequences as necessary, since $[0, 1]$ is compact, we may assume that $t_{n_k} \to t'$ for some $t' \in [0, 1]$. Note that $|t' - t| \ge \varepsilon$, and so $t \neq t'$.
Since $g$ is continuous, $g(t_{n_k}) = x_{n_k} \to g(t')$. But, $x_{n_k}$ is a subsequence of the convergent sequence $x_n \to x$, and so $x_{n_k} \to x$. This implies that $x = g(t') \implies t' = g^{-1}(x) = t$. This contradicts $t' \neq t$. Thus, $g$ is continuous.
From here we see, as Greg suggests, that $h:= g^{-1} \circ f$ is a continuous, bijective map from $[0, 1]$ to $[0, 1]$.
Consider $t_0 := h^{-1}(0)$. If $t_0 \in (0, 1)$, then $t_1:= t_0 / 2 \in (0, t_0)$ and $t_2 := (1 - t_0) / 2 \in (t_0, 1)$. By injectivity, $h(t_1), h(t_2) > 0$. Pick some $s \in (0, \min\{h(t_1), h(t_2)\})$. Then $0 < s < h(t_1)$ and $0 < s < h(t_2)$. By applying intermediate value theorem twice, we can find $t_3 \in (t_1, t_0) \subseteq (0, t_0)$ and $t_4 \in (t_0, t_2) \subseteq (t_0, 1)$ such that
$$h(t_3) = h(t_4) = s,$$
despite $t_3 < t_0 < t_4$. This contradicts the intermediate value theorem. Thus, $h^{-1}(0) \in \{0, 1\}$. A similar proof shows that $h^{-1}(1) \in \{0, 1\}$ too. Injectivity of $h^{-1}$ implies that these points are different, so
$$\{h^{-1}(0), h^{-1}(1)\} = \{0, 1\}.$$
Applying the bijection $g \circ h = f$ to both sets, we get
$$\{g(0), g(1)\} = \{f(0), f(1)\}.$$
