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Consider the function $f: [0,1) \rightarrow \mathbb{C}$ given by $f(t) = e^{2\pi i t}$. I must show that the function $f^*: [0,1) \rightarrow \mathrm{im}(f)$ is not a homeomorphism, given the standard topologies on both sets, but I am not sure how to proceed.

Some work: I am given that $f$ is injective (from which it follows that $f^*$ is surjective, and thus bijective and invertible). I also know that $f^*$ is continuous. Thus I need to show that the inverse of $f^*$ is not continuous. I'm unsure of how to compute the inverse directly (does this require any complex analysis? I have no background in it!), so I am trying to show that $f^*$ is not an open map by looking for an open set in $[0,1)$ that does not get mapped to an open set in $\mathrm{im}(f) = S^1$, with respect to the subspace topology induced by $\mathbb{C} \cong \mathbb{R}^2$.

Any help is appreciated!

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  • $\begingroup$ There is a point you can remove from $[0,1)$ which doesn't disconnect it--is this true of $S^1$? $S^1$ is compact, is $[0,1)$? $\endgroup$ Oct 10 '13 at 4:31
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Let $f^*:[0,1) \to \mathbb{S}^1$ be the map $f^*(t)=e^{2 i \pi t}$. Now define the the function $g:\mathbb{S} \to [0,1)$ as the mapping with the rule: $$ e^{2\pi i t } \mapsto t $$ It will be easy for you to show that this $g$ is the inverse of $f^*$. Now you should know that a function is continuous iff the inverse image of every open set is open. So consider the inverse image of $[0,\epsilon)$ under $g$ where $\epsilon<1$. The set $[0,\epsilon)$ is an open subset in the relative topology of $[0,1)$. But the inverse image is $$ g^{-1}([0,\epsilon))=\left\{ e^{2\pi i t} \mid t\in [0,\epsilon) \right\} $$ which is not open in $\mathbb{S}^1$. This is easy to verify since there is no open set containing $1=e^{2 i \pi \cdot 0}\in \mathbb{S}^1$ entirely contained in $g^{-1}([0,\epsilon))$. Every such open set will contain an element $e^{2 i \pi s}$ with $s<0$.

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Consider the subset $[0,\frac{1}{2})$. This is open in $[0,1)$, but is $f[0,\frac{1}{2})$ open in $\text{im}(f)$?

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Hint What is $\lim_{t \to 0^+} f(t)$? $\lim_{t \to 1^-} f(t)$?

What does this tell you about $\lim_{z \to 1} f^*(z)$?

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