When we say that $\mathbb S^1$ is compact, which topology we are thinking about?

Question

Let $$f:[0,2\pi)\to \mathbb S^1$$ defined as $$f(t)=e^{it}.$$ Then $f$ is continuous and bijective, but the inverse is not continuous because $f^{-1}(\mathbb S^1)=[0,2\pi)$ and $\mathbb S^1$ is compact whereas $[0,2\pi[$ it's not.

Q1) What is the definition of $\mathbb S^1$ ? Is it $\{e^{it}\mid t\in [0,2\pi]\}$ ? Is it $\{(\cos \theta ,\sin\theta )\mid \theta \in [0,2\pi]\}$ ? Or is it the quotient space $[0,1]/_\sim$ where $0\sim 1$ and $x\sim x$ for all $x\in (0,1)$ ?

Q2) So when we say it's compact, it's wrt which topology ? quotient ? metric ? btw what would be a metric on $\mathbb S^1$ ?

Im confused...

Answer 1

1. If you want that the definition of $S^1$ makes sense, define it as$$\left\{e^{it}\,\middle|\,t\in[0,2\pi]\,\right\}$$(or $t\in[0,2\pi)$ or $t\in\mathbb R$ or …).
2. The topology induced by the usual distance in $\mathbb C$:$$d(z,w)=\lvert z-w\rvert.$$