Two ordered points in the plane: centre and difference coordinates

$\mathrm{\Phi }$ is well defined. Let $(z_1,z_2)\in F_2(\mathbb{C})$, so $z_1\mathrm{\ne }{z}_2$. If $z_2-z_1=0$ then adding $z_1$ gives $z_2=z_1$ by [F2], a contradiction; hence $z_2-z_1\mathrm{\ne }0$ and $\mathrm{\Phi }(z_1,z_2)\in \mathbb{C}\times {\mathbb{C}}^{\times }$.

$\mathrm{\Psi }$ is well defined. Let $(c,w)\in \mathbb{C}\times {\mathbb{C}}^{\times }$, so $w\mathrm{\ne }0$. The two coordinates of $\mathrm{\Psi }(c,w)$ differ by $(c+\frac{w}{2})-(c-\frac{w}{2})=w\mathrm{\ne }0$, hence are distinct, and $\mathrm{\Psi }(c,w)\in F_2(\mathbb{C})$.

Linear combinations on the plane are continuous. Let $\alpha ,\beta \in \mathbb{C}$ and consider $L:{\mathbb{C}}^2\to \mathbb{C}$, $L(z_1,z_2):=\alpha z_1+\beta z_2$. Let $x=(x_1,x_2)\in {\mathbb{C}}^2$ and $\epsilon >0$, and put $\delta :=\epsilon \mathrm{/}(1+\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })>0$, a positive real. If $\mathrm{\mid }{z}_k-x_k\mathrm{\mid }<\delta$ for $k=1,2$, that is, if $z$ lies in the basic open box $B(x_1,\delta )\times B(x_2,\delta )$ of [L4], then by [L3] $\mathrm{\mid }L(z)-L(x)\mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }{z}_1-x_1\mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }\mathrm{\mid }{z}_2-x_2\mathrm{\mid }<(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\delta \le \epsilon ,$ with strict inequality $\mathrm{\mid }L(z)-L(x)\mathrm{\mid }<\epsilon$ in every case: if $\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }=0$ then $L(z)-L(x)=0<\epsilon$, and otherwise $(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\delta =\epsilon (\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1+\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })<\epsilon$. Hence the preimage of the ball $B(L(x),\epsilon )$ contains the box $B(x_1,\delta )\times B(x_2,\delta )$ around $x$, and since balls form a basis of the topology of $\mathbb{C}$ and boxes a basis of the topology of ${\mathbb{C}}^2$ [L4], $L$ is continuous by [L5].

$\mathrm{\Phi }$ and $\mathrm{\Psi }$ are mutually inverse, so $\mathrm{\Phi }$ is a bijection. Let $(z_1,z_2)\in F_2(\mathbb{C})$ and put $c:=\frac{z_1+z_2}{2}$, $w:=z_2-z_1$. By the field laws of [F2] and $2\cdot \frac{1}{2}=1$, $c-\frac{w}{2}=\frac{z_1+z_2-(z_2-z_1)}{2}=\frac{2z_1}{2}=z_1,c+\frac{w}{2}=\frac{z_1+z_2+(z_2-z_1)}{2}=\frac{2z_2}{2}=z_2,$ so $\mathrm{\Psi }(\mathrm{\Phi }(z_1,z_2))=(z_1,z_2)$. Conversely, for $(c,w)\in \mathbb{C}\times {\mathbb{C}}^{\times }$ the first coordinate of $\mathrm{\Phi }(\mathrm{\Psi }(c,w))$ is $\frac{(c-\frac{w}{2})+(c+\frac{w}{2})}{2}=c$ and the second is $(c+\frac{w}{2})-(c-\frac{w}{2})=w$, so $\mathrm{\Phi }(\mathrm{\Psi }(c,w))=(c,w)$. Thus $\mathrm{\Phi }$ has the two-sided inverse $\mathrm{\Psi }$ and is a bijection.

$\mathrm{\Phi }$ is continuous. The two components of $\mathrm{\Phi }$ are the restrictions to the subspace $F_2(\mathbb{C})\subseteq {\mathbb{C}}^2$ of the continuous maps $L_{\frac{1}{2},\frac{1}{2}}(z_1,z_2)=\frac{z_1+z_2}{2}$ and $L_{-1,1}(z_1,z_2)=z_2-z_1$ of step 1.3, hence are continuous by [F1]. The second of them takes all its values in the subspace ${\mathbb{C}}^{\times }\subseteq \mathbb{C}$ by step 1.1, so it is a continuous map $F_2(\mathbb{C})\to {\mathbb{C}}^{\times }$ by the subspace criterion of [F1]; therefore $\mathrm{\Phi }$, whose target is the product $\mathbb{C}\times {\mathbb{C}}^{\times }$, is continuous by the product criterion of [L5].

$\mathrm{\Psi }$ is continuous. The target $F_2(\mathbb{C})$ is a subspace of ${\mathbb{C}}^2$, so by the subspace criterion of [F1] it suffices to show that the composite $H:\mathbb{C}\times {\mathbb{C}}^{\times }\to {\mathbb{C}}^2$, $H(c,w)=(c-\frac{w}{2},c+\frac{w}{2})$, is continuous. By the product criterion of [L5] it suffices that the two components $L_{1,-\frac{1}{2}}(c,w)=c-\frac{w}{2}$ and $L_{1,\frac{1}{2}}(c,w)=c+\frac{w}{2}$ be continuous as maps into $\mathbb{C}$. The domain $\mathbb{C}\times {\mathbb{C}}^{\times }$ is a subspace of ${\mathbb{C}}^2$: by [L4] its basic open sets are the boxes $U\times (V\cap {\mathbb{C}}^{\times })$ with $U,V$ open in $\mathbb{C}$, and these are exactly the traces $(U\times V)\cap (\mathbb{C}\times {\mathbb{C}}^{\times })$ of the boxes $U\times V$ on ${\mathbb{C}}^2$, so the two topologies coincide. Hence continuity of the components follows from the continuity of $L_{1,\mp \frac{1}{2}}$ on ${\mathbb{C}}^2$ in step 1.3 together with the restriction clause of [F1].

Conclusion. By step 2.1 and step 1.2 the map $\mathrm{\Phi }$ is a bijection $F_2(\mathbb{C})\to \mathbb{C}\times {\mathbb{C}}^{\times }$ with inverse $\mathrm{\Psi }$; by steps 2.2 and 2.3 both $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are continuous. Hence $\mathrm{\Phi }$ is a homeomorphism and $F_2(\mathbb{C})\cong \mathbb{C}\times {\mathbb{C}}^{\times }$, as claimed. No choice principle was used.