Finite join models for the circle and the two-point group

The nonnegative square-root function used below is continuous. Indeed, for $u,v\ge 0$, assume without loss that $u\ge v$. Since $v^2\le uv$, nonnegativity and uniqueness in [F2] give $v\le \sqrt{uv}=\sqrt{u}\sqrt{v}$. Hence [F2, algebra] $(\sqrt{u}-\sqrt{v}{)}^2=u+v-2\sqrt{uv}\le u-v,$ so $\mathrm{\mid }\sqrt{u}-\sqrt{v}\mathrm{\mid }\le \sqrt{\mathrm{\mid }u-v\mathrm{\mid }}$. Given $\epsilon >0$, taking $\mathrm{\mid }u-v\mathrm{\mid }<{\epsilon }^2$ proves continuity, including at zero.

On the quotient presentation in [F1], define [F1, F5, step 1.1] ${\mathrm{\Phi }}_N\text{\hspace{0.17em}⁣}\left({\sum }_{i=0}^N{t}_i{z}_i\right)=(\sqrt{t_0}{z}_0,\dots ,\sqrt{t_N}{z}_N)\in {\mathbb{C}}^{N+1}\mathrm{.}$ Its squared norm is ${\sum }_i{t}_i=1$. The formula is independent of every $z_i$ with $t_i=0$, and its formula before quotienting is continuous by [F5] and step 1.1. It therefore descends to a continuous map $(S^1{)}^{\ast (N+1)}\to S^{2N+1}$. It is onto: for $w=(w_i)$ on the unit sphere use $t_i=\mathrm{\mid }{w}_i{\mathrm{\mid }}^2$ and, when $w_i\mathrm{\ne }0$, $z_i=w_i\mathrm{/}\mathrm{\mid }{w}_i\mathrm{\mid }$; labels at zero coordinates may be set to $1$. It is injective, because its image recovers every $t_i=\mathrm{\mid }{w}_i{\mathrm{\mid }}^2$ and every label at a positive weight, which is exactly the equivalence relation in [F1].

Similarly define [F1, F5, step 1.1] ${\mathrm{\Psi }}_N\text{\hspace{0.17em}⁣}\left({\sum }_{i=0}^N{t}_i{\epsilon }_i\right)=({\epsilon }_0\sqrt{t_0},\dots ,{\epsilon }_N\sqrt{t_N})\mathrm{.}$ It is well defined and continuous by the same argument as step 2.1. For a point $x=(x_i)\in S^N$, recover $t_i=x_i^2$ and, at a positive weight, ${\epsilon }_i$ as the sign of $x_i$. This proves bijectivity, because the recovered data agree exactly at every positive weight.

The simplex, the circle, the finite subset $\{1,-1\}$, and both target spheres are closed bounded subsets of finite-dimensional Euclidean spaces, hence compact by [F3]; the relevant finite products remain compact. Each quotient source is a continuous image of its compact product and is compact by [F4], while each target sphere is Hausdorff by [F6]. Thus the continuous bijections in steps 2.1--2.2 are homeomorphisms by [F4]. This also shows that the ordinary compact quotients are already compactly generated, so they agree with the standing kified finite-join convention.

Appending a zero weight appends the zero target coordinate in both formulas, so the homeomorphisms commute with the standard inclusions. For $\lambda \in S^1$, [F1, step 3.1] ${\mathrm{\Phi }}_N\text{\hspace{0.17em}⁣}\left(\left({\sum }_i{t}_i{z}_i\right)\lambda \right)=(\sqrt{t_i}{z}_i\lambda {)}_i={\mathrm{\Phi }}_N\text{\hspace{0.17em}⁣}\left({\sum }_i{t}_i{z}_i\right)\lambda \mathrm{.}$ Thus the first map is equivariant. Its orbit quotient is the unit-sphere quotient by phases, which is ${\mathbb{C}\mathbb{P}}^N$: every nonzero complex vector has a unique positive radial normalization, and two unit vectors span the same complex line exactly when they differ by a unit phase. Likewise multiplication of every ${\epsilon }_i$ by $-1$ sends ${\mathrm{\Psi }}_N$ to its antipode, and the second orbit quotient is $S^N\mathrm{/}(x\sim -x)={\mathbb{R}\mathbb{P}}^N$.

At $N=0$, ${\mathrm{\Phi }}_0$ is the identity of $S^1$ and its orbit quotient is one point; ${\mathrm{\Psi }}_0$ identifies the two-element group with $S^0$ and its orbit quotient is one point. No coordinate with zero weight is ever divided by, and all products and label assignments are finite. Thus the endpoint and degenerate cases introduce no choice, and steps 1.1--4.1 prove every claim. $\mathrm{\square }$