Which of the following maps define a homeomorphism? Which of the following maps define a homeomorphism?
(a) $f : \mathbb R \to ]0,\infty[$, where $f(x) = e^x.$
(b) $f : [0, 1] \to S^1$, where $f(t) = (\cos 2\pi t, \sin 2\pi t).$
(c) Any map $f: X \to Y$ which is continuous, one-one and onto, if $X$ is
compact and $Y$ is Hausdorff.
My solution
(a) $f : \mathbb R \to ]0,1[$ is continuous. $f^{-1}=\log(x)$ is also continuous in $(0,1)$. So, $f : \mathbb R \to ]0,1[$ is homeomorphism.
(b) $f$ is not homeomorphism. since the removal of a single element from$[0,1]$ make it disconnected. But, from $S^1$ doesn't disturb the connectedness.
(c)Here, Given map is bijective, continuous map. I need only prove $f$ is an open maping inorder to prove $f$ is a homeomorphism.
 A: Your solution for (a) is incorrect because $e^x>1$ for $x>0$, so the codomain is not correct. 
Your solution for (b) is correct. One could also argue that $f(0)=f(1)$, so $f$ is not injective. If one defines $\sim$ on $[0,1]$ by setting $0 \sim 1$ and $x \sim x$ for all $x \in (0,1)$, then the induced map $[0,1]/_\sim \to S^1$ is indeed a homeomorphism, by (c).
Option (c) is correct and the only thing we need to show is that $f^{-1}$ is also continuous. As you said, it suffices to show that $f$ is an open map. But it is easier to show that $f$ is a closed map, as follows: let $C\subseteq X$ be closed. Since $X$ is compact, $C$ is compact. Then $f(C)$ is compact by continuity. Since $Y$ is Hausdorff, $f(C)$ is closed.
A: (a) is incorrect because $f(\mathbb R) \not\subset ]0,1[$, for example $f(1)=e^1 \approx 2.718... \not\in ]0,1[$.
For (b), you are right that it is not a homeomorphism, but there is a simpler reason unrelated to connectedness: $f$ is not one-to-one, because $f(0)=f(1)$.
For (c), your outline is correct, but there's still work to do.
