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A maximal external point of a finite set has no other point strictly larger in every coordinate. This weak nondominance differs from a maximal element in the product order when coordinate ties exist: for example $(1,0)$ passes the weak test against $(1,1)$ but is not maximal in the product order. In an independent continuous-coordinate random sample, ties have probability zero, so the two notions coincide almost surely.

For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe relations $jz=\overline{z}j$ and $j^2=-1$ verify multiplication, and every matrix in $SU(2)$ has this form. If $v$ is imaginary, then $\overline{v}=-v$. Quaternionic conjugation reverses product order and $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. Multiplicativity of the quaternion norm shows that this action preserves the inner product. The unit quaternions form the connected three-sphere, so its determinant, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.

Its group kernel consists of unit quaternions commuting with every imaginary quaternion. Commuting with both $i$ and $j$ forces a quaternion to be real, so the group kernel is exactly $\{\pm 1\}$. To prove surjectivity explicitly, let $u$ be a unit imaginary quaternion and put $q=\cos (\theta /2)+u\sin (\theta /2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qv{q}^{-1}$ therefore givesThis is the Rodrigues rotation formula about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional orthogonal matrix of determinant one has an eigenvector of eigenvalue one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.

Here a pointed complete partial order is a partially ordered set $D$ with a least element ${\perp }_D$ in which every nonempty directed set has a least upper bound. Directed means that every pair of its elements has an upper bound within that set. A Scott continuous map is monotone and preserves these directed suprema. It is not required to preserve the least element; restricting to strict maps would change the categorical claim. If completeness is instead formulated only for increasing countable chains, the same arguments below work with those chains.

Order the function space of complete partial orders $[D\to E]$ pointwise: $f\le g$ means $f(x)\le g(x)$ for every $x$. Its least element is the constant function with value ${\perp }_E$. For a directed family $\mathcal{F}$ of continuous maps, define $h(x)={\sup }_{f\in \mathcal{F}}f(x)$. This is well-defined and is its pointwise least upper bound. It is monotone. For any directed $A\subseteq D$,The interchange is valid because the combined family $\{f(a):f\in \mathcal{F},a\in A\}$ is directed, and either iterated supremum is its least upper bound. Thus $h$ is a Scott continuous map, proving that $[D\to E]$ is a pointed complete partial order.

Finite products carry the product order, componentwise suprema and least element $({\perp }_D,{\perp }_E)$. The one-element order is a terminal object. For the exponential, use the function space with evaluation $\mathrm{ev}(f,x)=f(x)$. To check joint continuity, take a directed family of pairs $(f_i,x_i)$. ThenIndeed, directedness supplies a $k$ above the pairs indexed by both $i$ and $j$, giving $f_i(x_j)\le f_k(x_k)$. Thus the diagonal values are cofinal in the double family, and evaluation preserves its supremum.

For a continuous $h:A\times D\to E$, its currying is $\Lambda h:A\to [D\to E]$, $(\Lambda h)(a)(d)=h(a,d)$. Each section is continuous because $d\mapsto (a,d)$ is; pointwise directed suprema in the $a$ coordinate show that $\Lambda h$ is continuous. Conversely a continuous $g:A\to [D\to E]$ gives the continuous map $\mathrm{ev}\circ (g\times {\mathrm{id}}_D)$. The two constructions are inverse, because they agree on every pair $(a,d)$. This gives the natural bijectionwith the required evaluation map. Together with the terminal object and products, it proves cartesian closure of pointed complete partial orders with non-strict continuous maps.