I am working with a topic of quotient map and homeomorphism. Assume $X = [0, 2\pi]$ and I define a map $f: X \rightarrow S^{1}$ by the function $$f(t) = e^{it}$$ a) If I assume $\sim$ is the equivalence relation on X, given by $$\forall x \in X, x \sim x. \text{Also}\, 0 \sim 2\pi, \text{and}\, 2\pi \sim 0.$$ I denote $q: X \rightarrow X/\sim$ be the quotient map. Show that there is a homeomorphism $\bar{f}: X/\sim \rightarrow S^{1}$ which satisfies $\bar{f} \circ q = f.$
I am wondering how to show such a map is a homeomorphism. Do I have to show that it is well-defined and then show that it is bijective to satisfy the definition of a homeomorphism?
b) Now, I want to denote another map g which is the restriction of f to $[0,2\pi]$. Would g be continuous and bijective? I am wondering since g is the restriction of f, it may somewhat have some properties from f, so g can be a homeomorphism as well. But I could be wrong since I cannot find a way to prove or disprove it.
Any tip would be helpful. Thank you in advance!
By the universal mapping property of quotient spaces, the fact that $f$ identifies $0$ and $2\pi$ means that your $\overline{f}$ exists and is continuous. It is also onto because $f$ is. It is one-to-one because $f$ was EXCEPT at $0$ and $2\pi$, but you glued these points together. Finally, the quotient is compact because $[0,2\pi]$ is and $S^1$ is Hausdorff, so $\overline{f}$ is a homeo.