How homeomorphic are noninjective images of $[0,1]$ to $[0,1]$? I have a continuous function $f:[0,1]\rightarrow \mathbb{R^2}$. If $f$ we injective, we'd know the image is homeomorphic to $[0,1]$. Lets consider $S:=\{f(t) : f \text{ injective at }  
t\}$ (i.e $t$ such that $f^{-1}\big(\{f(t)\}\big)=\{t\}$ .)
What can we say about the homeomorphism type of S as a subspace of $\mathbb{R^2}$? Is $S$ homeomorphic to a subset of $[0,1]$? In particular, if $S$ is dense in the image of $f$, must $S$ be homeomorphic to a dense subset of $[0,1]$?
Clearly the same theorem doesnt apply since $S$ need not be compact. We could choose big compact subsets of $S$ and show that almost all of $S$ is homeomorphic to some union of compact intervals, therefore embeds in $[0,1]$ nicely.
 A: Let $f:X\rightarrow Y$ be continuous, with $X,Y$ metric spaces and $X$ compact, and let $$A=\{t\in X:f(t')=f(t)\rightarrow t'=t\}\quad \quad\quad S=f(A)$$
Your case is with $X=[0,1]$, $Y=\mathbb{R}^2$.
Now clearly, the restriction $f\rvert_A:A\rightarrow S$ is a continuous bijection, so we need only show that $(f\rvert_A)^{-1}:S\rightarrow A$ is continuous. To that end, let $p\in A$, and let $(a_n)$ be a sequence of points in $A$ such that $(f(a_n))$ converges to $f(p)$. Now suppose $(a_n)$ does not converge to $p$. Then there is some neighborhood $N$ of $p$, such that $a_n\not\in N$ for infinitely many $n$. Let $(b_n)$ be a subsequence of $(a_n)$ with $b_n\not\in N$ for each $n$. Then $X-N$ is compact, so $(b_n)$ has a subsequence $(c_n)$ that converges to some point $c\in X-N$. Since $f:X\rightarrow Y$ is continuous,
$$\lim_{n\rightarrow \infty} f(c_n)=f(c)$$
However, $(c_n)$ is a subsequence of $(a_n)$, so
$$\lim_{n\rightarrow \infty} f(c_n)=\lim_{n\rightarrow \infty} f(a_n)=f(p)$$
which is a contradiction, since $c\neq p$. We conclude that $(a_n)$ must in fact converge to $p$, which proves that $(f\rvert_A)^{-1}:S\rightarrow A$ is continuous. Thus, $f\rvert_A:A\rightarrow S$ is a homeomorphism.
Now it's pretty clear that if $X=[0,1]$ and $S$ is dense in $\text{Im}(f)$, $A$ need not be dense in $[0,1]$, for if we have a function $f$ with $A(f)$ dense in $[0,1]$, then
$$h(t)=\begin{cases}
f(2t) & 0\leq t\leq 1/2\\ 
f(1) & 1/2\leq t\leq 1
\end{cases}$$
has $S(f)-\{f(1)\}=S(h)$, but $A(h)\subseteq [0,1/2]$, which is not dense in $[0,1]$.
