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?

  • 1
    $\begingroup$ To answer your aside question, $f,g$ continuous and $f=g\circ h$ doesn't mean $h$ is continuous. As a simple example, let $h$ be any noncontinuous function and let $g$ be the constant map. $\endgroup$ – Cheerful Parsnip Dec 22 '18 at 19:52

$\mathbb{R}P^2$ 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 $S^2_+$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.