Let $\mathbb{S}^2$ be the unit sphere in $\mathbb{R}^3$. Let $f:\mathbb{S}^2\rightarrow\mathbb{R}^4$ be defined by $f(x,y,z)=(x^2-y^2,xy,yz,zx)$
prove that $f$ determines a continuous map $\tilde{f} : \mathbb{R}P^2 → \mathbb{R}^4$ where $\mathbb{R}P^2$ is the real projective plane and that $\tilde{f}$ defines a homeomorphism onto a subset of $\mathbb{R}^4$
$\mathbb{R}P^2$ is homeomorphic to the upper hemisphere of the sphere so is the first proof to show that $f$ is continuous on the upper hemisphere and that we can map the sphere continuously onto the upper hemisphere?
[as an aside question if $f = g\cdot h$ and $f,g$ are continuous, is $h$ continuous?]
Fir the second part, is the subset it is homeomorphic to just the image of the upper hemisphere?
Best Answer
$\mathbb{R}P^2$ is not homeomorphic to the upper hemisphere. In fact, it is the quotient of $S^2$ obtained by identifying all antipodal points $p,-p$.
Hence to see that $f$ determines $\tilde{f}$, it suffices to verify that $f(p) = f(-p)$. But this is obvious from the definition.
To prove that $\tilde{f}$ is an embedding, it suffices to show that it is injective (use the compactness of $\mathbb{R}P^2$). In other words, we have to show that $f(x,y,z) = f(x',y',z')$ implies $(x',y',z') = \pm (x,y,z)$. So let $$(*) \phantom{xx} (x^2-y^2,xy,yz,zx) = ((x')^2-(y')^2,x'y',y'z',z'x') .$$
Write $c = x + iy \in \mathbb{C}$, $c' = x' + iy' \in \mathbb{C}$. Then $c^2 = x^2 - y^2 + i(2xy)$, $(c')^2 = (x')^2 - (y')^2 + i(2x'y')$. $(*)$ shows that $c^2 = (c')^2$, i.e. $c'= \epsilon c$ with $\epsilon = \pm 1$. In other words, $(x',y') = \epsilon (x,y)$.
Case 1. $(x,y) = (0,0)$. Then $z,z' \in \{ -1, 1\}$, hence $(x',y',z') = \pm (x,y,z)$.
Case 2. $(x,y) \ne (0,0)$. W.lo.g. let $x \ne 0$. Then $\epsilon x z' = x'z' = xz$, hence $z' = \epsilon z$. This shows $(x',y',z') = \epsilon (x,y,z)$.
Concerning your final question: The image $R = \tilde{f}(\mathbb{R}P^2)$ agrees with the image $f(S^2)$. And in fact $f(S^2) = f(S^2_+)$, where $S^2_+$ is the upper hemisphere of $S^2$. But note that $f \mid_{S^2_+}$ is no embedding because $f \mid_{S^2_+}$ still identifies antipodal points in the equator $E$ of $S^2$ which is contained in $S^2_+$.