# Question on proving quotient space is homeomorphic to circle [duplicate]

I am new to quotient spaces and was given this in class. I really have no idea how to solve. I tried one approach that my teacher said to be incorrect so I'd really appreciate the help on this.

I am given $$S^1= \{x \in \mathbb{R}^2 : \|x\|=1 \}$$ i.e., the unit circle in $\mathbb{R}^2$, and I am asked to show the quotient space $S^1 / {\sim}$, where $\sim$ is the equivalence relation $x\sim{-x}$, is homeomorphic to $S^1$.

My attempt was to look at the quotient space as the set of points of the unit circle with only one of $(1,0)$ and $(0,1)$, which I now realize is incorrect as I know quotient space means gluing as opposed to omitting points. I am stumped as I cannot really figure out how to show an explicit homeomorphism between the quotient space and the unit circle in $\mathbb{R}^2$, or if I am even on the right track (maybe I need to use a theorem or something without a constructive proof). I really would appreciate the help on this. Thanks all.

## marked as duplicate by Zev Chonoles, Adam Hughes, drhab, Alex S, user147263 Jul 8 '15 at 18:01

Label each point on the circle by the angle $\theta\in [0,2\pi)$. Let $f:S/{\sim}\rightarrow S$ be defined by $f(\theta)=2\theta$. You should be able to prove this is well-defined on $S/{\sim}$, and is in fact a bijection between the two spaces. Then to prove it's a homeomorphism, you just need to show that the image/preimage of any open set is open.