Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1\subseteq \mathbb{R}^2$.
My first idea was to define $g\colon I\to S^1$ by $x\mapsto (\cos(2\pi x), \sin(2\pi x))$, but then the map $f$ satisfying $g=f\circ p$, where $p$ is the quotient map, seems not to be injective. Any hints?
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I'm pretty sure you can show that $I=[-1,1] \approx [0,1]$, say by $h : [-1,1] \to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/{\sim}$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X \to Y$ which makes same identification as projection map $p : X \to X/{\sim}$. In your case define $q : [0,1] \to \Bbb{S}^1$ by $s \mapsto e^{2\pi is}$. $\require{AMScd}$ \begin{CD} I @>{h}>> [0,1]\\ @VpVV @VVqV\\ I/{\sim} @>>> \Bbb{S}^1 \end{CD} Since $q \circ h$ is a quotient map that makes same identification as $p$, then $I/{\sim} \approx \Bbb{S}^1$ by Uniqueness of Quotient Topology.
$$x \mapsto (\cos(\pi x), \sin(\pi x))$$