Is every continuous one-to-one image of $[0,\infty)$ locally compact?

Question

Suppose $f:[0,\infty)\to Y$ is continuous and one-to-one onto $Y$. You may assume $Y$ is metric.

Is $Y$ locally compact?

Thanks!

Answer 1

No, here is a counterexample. Define $f:[0,\infty)\to\mathbb R\times\mathbb R$ as follows: $$f(t)= \begin{cases} (t,\,-t)\text{ if }0\le t\le\frac12,\\ (t,\,t-1)\text{ if }\frac12\le t\le1,\\ (\frac1t,\,\sin^2\pi t)\text{ if }1\le t\lt\infty.\\ \end{cases}$$ The function $f$ is continuous and one-to-one. The set $S=\operatorname{range}f$ is not locally compact, because the point $(0,0)=f(0)\in S$ has no compact neighborhood in $S$. (Note that, if $0\lt\varepsilon\lt1$, then the intersection of $S$ with the horizontal line $y=\varepsilon$ has a limit point $(0,\varepsilon)\notin S$.)