Show that the real projective plane $\mathbb{R}P^2$ is homeomorphic to $S^2/ \langle x \sim -x \rangle$.
I know that $S^2/ \langle x \sim -x \rangle$ is the quotient space of the quotient map $g:S^2 \rightarrow S^2/\langle x \sim -x \rangle$, given by $g(x)=\{x,-x\}$ for all $x \in S^2$. So to construct a homeomorphism $f:\langle x \sim -x \rangle \rightarrow \mathbb{R}P^2$, I would define $f(\{x,-x\})=tx$. Where $t$ is a parameter.
Then similarly to the last homeomorphism problem I did, I want to show that $f$ is a continuous bijection from a compact space to a Hausdorff space. I know $S^2/\langle x \sim -x \rangle$ is compact since it is the surjective image of the continuous quotient map $g$ and $S^2$ is compact. I know $\mathbb{R}P^2$ is Hausdorff, but the proof seems difficult.
$f$ is injective since $f(\{x,-x\})=f(\{y,-y\})\implies tx=ty \implies \{x,-x\}=\{y,-y\}$.
$f$ is surjective, since given any $tx \in \mathbb{R}P^2$, $\tilde{x}=\frac{x}{||x||} \in S^2$, so that $f(\{\tilde{x},-\tilde{x}\})=tx$.
Then all that is left to show is that $f$ is continuous (after showing that $\mathbb{R}P^2$ is Hausdorff), how should I go about showing that $f$ is continuous? Normally I would use a theorem and the projection functions to do this, but I know that the projective plane cannot be embedded into $\mathbb{R}^3$, so I do not think this would suffice.
I want to use the following to show continuity:if $q: X \to Y$ is a quotient map then $g: Y \to Z$ is continuous iff $g \circ q: X \to Z$ is continuous. So continuity from a quotient space is determined by the composition with the quotient map.
I tried to show the composition mapping $f \circ g:S^2 \rightarrow \mathbb{R}P^2$ is continuous, but this map is given by $(f\circ g)(x)=tx$ so I do not know how continuity would follow from that.
Also I know that a basis for $\mathbb{R}P^2$ is given by the collection of subsets of lines specified by open double cones with the cone point at the origin, so I think it would be difficult to show if $U$ is open in $\mathbb{R}P^2$ then $f^{-1}(U)$ is open in $S^2/\langle x \sim -x \rangle$. Any ideas?
