Proving constructed homeomorphism map is continuous I was constructing a homeomorphism from $S^2 / \langle x \sim -x \rangle$ to $\mathbb{R}P^2$, the real projective plane, defined to be the space of all lines through the origin in $\mathbb{R}^3$.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 .
The homeomorphism I came up with is $f:S^2 / \langle x \sim -x \rangle \rightarrow \mathbb{R}P^2,$ given by $f(\{x,-x\})=tx$.Where $t$ is a parameter.
I was having some difficulty showing that this map is continuous and came across the following theorem:
Let $p:X \rightarrow Y$ be a quotient map. Let $Z$ be a space and let $g:X \rightarrow Z$ be a map that is constant on each set $p^{-1}(\{y\})$, for $y \in Y$. Then $g$ induces a map $f:Y \rightarrow Z$ such that $f \circ p=g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Can I use this theorem to show the function $f$ I have constructed is continuous? For this I will denote $p:S^2 \rightarrow S^2/ \langle x \sim -x \rangle$ to be the quotient map, sending points $x,-x$ on the sphere to the equivalence class $\{x,-x\}$. So then in terms of the theorem I can take $p=p,X=S^2, Y=S^2/ \langle x \sim -x \rangle,Z=\mathbb{R}P^2,f=f,g=g$ and $g:S^2 \rightarrow \mathbb{R}P^2$ given by $g(x)=tx$ for all points in the sphere. Then $g$ is constant on $p^{-1}(\{\{x,-x\}\})$, since $g(x)=tx,g(-x)=tx$.
Can I just say $g$ is continuous since each of the components of $tx$ is a continuous polynomial, or is that not allowed since $\mathbb{R}P^2$ cannot be embedded in $\mathbb{R}^3$?
Now I am unsure whether this makes any progress, because I still have to show $g$ is continuous, which seems nonobvious. Any help?
 A: To show $g$ is continuous, you have to show that for any basis element $B$ of $\mathbb RP^2$, $g^{-1}(B)$ is an open subset of $S^2$.
As you say, $B$ is a collection of lines through the origin whose union is an open cone $C_B \subset \mathbb R^3$. Working through the definitions we get
$$g^{-1}(B) = S^2 \cap C_B
$$
Now there is a bit of trouble because $C_B$ is not an open subset of $\mathbb R^3$. However, what it means to say that $C_B$ is an "open cone" is precisely that $C_B - \{0\}$ is an open subset of $\mathbb R^3 - \{0\}$. And since $0 \not\in S^2$ it follows that $S^2$ is a subspace of $\mathbb R^3 - \{0\}$, therefore $S^2 \cap C_B$ is an open subset of $S^2$, finishing the proof of continuity of $g$.
A: We know that $\Bbb R P^n : = \Bbb R^{n+1} - \{0\}/ \Bbb R - \{0\}.$ Now we have a surjective continuous map $f$ from $\Bbb R^{n+1} - \{0\}$ onto $\Bbb S^{n}$ defined by $$x \mapsto \frac {x} {\|x\|},\ x \in \Bbb R^{n+1} - \{0\}.$$ Let $p : \Bbb S^{n} \longrightarrow \Bbb S^{n}/ x \sim -x$ be the quotient map. Then the map $g : = p \circ f : \Bbb R^{n+1} - \{0\} \longrightarrow \Bbb S^n/ x \sim -x$ gives us a surjective continuous map. Let $X^* = \left \{g^{-1} \{z\}\ |\ z \in \Bbb S^n/ x \sim -x \right \}$ and endow $X^*$ with the quotient topology. Then by universal property of quotient topology $g$ induces a bijective continuous map $\overline {g} : X^* \longrightarrow \Bbb S^n /x \sim -x.$ Furthermore, $\overline {g}$ is a homeomorphism if and only if $g$ is a quotient map.
So if we can able to show that $X^* \approx \Bbb R^{n+1} - \{0\} / \Bbb R - \{0\}$ and $g$ is a quotient map then we are through by universal property of quotient topology.
For any $z \in \Bbb S^{n}.$ Now let $w \in g^{-1} \{[z]\},$ where $[z]$ is the equivalence class of $z \in \Bbb S^{n}$ in $\Bbb S^{n} / x \sim -x.$ Then $$\frac {w} {\|w\|} = \pm z \implies w = \pm \|w\| z \in \{\lambda z\ |\ \lambda \in \Bbb R - \{0\}\}.$$  Conversely, if $\lambda \in \Bbb R - \{0\}$ then $$g(\lambda z) = \left [ \frac {\lambda} {\|\lambda\|} z \right ] = [\lambda' z] = [z]$$ since $\lambda' = \frac {\lambda} {\|\lambda\|} = \pm 1.$ Therefore $g^{-1} \{[z]\} = \{\lambda z\ |\ \lambda \in \Bbb R - \{0\} \},$ which is nothing but the orbit of $z \in \Bbb R^{n+1} - \{0\}$ under the action of $\Bbb R - \{0\}$ on $\Bbb R^{n+1} - \{0\}$ which is given as follows $:$
For any given $z \in \Bbb R^{n+1} - 0$ and for any $z \in \Bbb R - \{0\}$ we define $$z \cdot (z_0,z_1, \cdots, z_n) : = (zz_0, zz_1, \cdots, zz_n).$$
So we have $$X^* = \Bbb R^{n+1} - \{0\} / \Bbb R - \{0\}.$$
Also since $f$ is a retraction it is a quotient map and thus $g = p \circ f,$ is a quotient map being the composition of two quotient maps. Thus we are the done with the proof!
