# Continuous mapping between $\mathbb{S}^2$, $\mathbb{R}P^2$ and $\mathbb{R}^4$

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?

• 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. – 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 $$S^2_+$$.