Homeomorphism between $[-\pi,\pi]$ and $S^1.$ Are $[-\pi,\pi]$ and $S^1$ homeomorphic? If so, is there any way to get an explicit homeomorphism? I am thinking about the map $t \mapsto e^{it}.$ But the problem is that this map fails to be $1$-$1$ at the endpoints $-\pi$ and $\pi.$ Is there any other clever way of doing this?
Any help in this regard would be much appreciated. Thanks for your time.
 A: HINT. Does removing a point from each set leave them connected? Is connectedness preserved under homeomorphism?
Any claims you make about each of these questions you should prove!
A: Try to remove one point to each one of your object. Can you note anything interesting? What is the other basic concept often introduced with compactness?
Edit: By the way, the question is fairly standard, I'm sure you can find the answer with the search function on math.stackexchange
A: There have already been some hints why $[-\pi,\pi]$ and $S^1$ might not be homeomorphic.
However, by identifying the endpoints of $[-\pi,\pi]$, you'll actually obtain $S^1$ (up to homeomorphism).
Identify (glue) the two endpoints $-\pi$ and $\pi$ of your line $[-\pi,\pi]$, from which you'll obtain the quotient space $$[-\pi,\pi]\big/{\sim}$$
And this is in fact homoemorphic to $S^1$. To see this, let $\exp\colon [-\pi,\pi] \to S^1$ be the map $\exp(t) = e^{it}$ and $p\colon [-\pi,\pi]\to [-\pi,\pi]\big/{\sim}$ the canonical projection.
By construction, we have $p(-\pi) = p(\pi) = \overline{\pi}$.
Furthermore, we have $$\exp(-\pi) = e^{i\pi} = e^{i(-\pi)} = \exp(\pi)$$
Thus, we descend to the quotient and $\exp\colon [-\pi,\pi] \to S^1$ induces a map
$$\varphi\colon [-\pi,\pi]\big/{\sim} \to S^1$$
defined via $$\varphi([x]) := f(x)$$  where $[x]$ denotes the equivalence class $[x]$ in your quotient space for $x \in [-\pi,\pi]$.
Now $\varphi$ is a homeomorphism since $[-\pi,\pi]$ is compact and $S^1$ as a subspace of $\mathbb{R}^2$ is Hausdorff, thus:
$$[-\pi,\pi]\big/{\sim} \cong S^1$$
