[proofplan] This proof is a sketch in the sense that the surjectivity step relies on a generation result for $\operatorname{SO}(3)$ (rotations in coordinate planes generate the group) that we cite rather than reprove from scratch. All other steps are fully rigorous. The strategy: realise $\operatorname{SU}(2)$ as the group of unit quaternions, identify the imaginary quaternions $\mathbb{H}^0$ with three-dimensional Euclidean space (where the Euclidean norm is the determinant), and show that conjugation $A \mapsto X A X^{-1}$ by $X \in \operatorname{SU}(2)$ is a norm-preserving linear map on $\mathbb{H}^0$. This gives a continuous group homomorphism $\varphi: \operatorname{SU}(2) \to \operatorname{O}(3)$ with kernel $\{\pm I\}$. Connectedness of $\operatorname{SU}(2)$ forces the image into $\operatorname{SO}(3)$, and explicit computation of $\varphi$ on three families of one-parameter subgroups exhibits rotations of arbitrary angles in three coordinate planes, generating $\operatorname{SO}(3)$. The induced bijection $\bar{\varphi}: \operatorname{SU}(2)/\{\pm I\} \to \operatorname{SO}(3)$ is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism, and the homeomorphism respects group structure by construction. [/proofplan] [step:Identify the imaginary quaternions $\mathbb{H}^0$ with three-dimensional Euclidean space] Let $\mathbb{H} = \{A \in \operatorname{Mat}_2(\mathbb{C}) : A = \alpha I + \beta i + \gamma j + \delta k, \, \alpha, \beta, \gamma, \delta \in \mathbb{R}\}$ be the algebra of quaternions, realised as $2 \times 2$ complex matrices with the basis \begin{align*} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \, i = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, \, j = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \, k = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}. \end{align*} A direct computation verifies the quaternion relations $i^2 = j^2 = k^2 = -I$, $ij = k$, $jk = i$, $ki = j$. The space of pure (imaginary) quaternions is \begin{align*} \mathbb{H}^0 := \{A \in \mathbb{H} : \operatorname{tr}\, A = 0\} = \mathrm{span}_\mathbb{R}(i, j, k), \end{align*} a three-dimensional real vector space. In coordinates $A = \beta i + \gamma j + \delta k$, the matrix is \begin{align*} A = \begin{pmatrix} i\beta & \gamma + i\delta \\ -\gamma + i\delta & -i\beta \end{pmatrix}. \end{align*} *Norm.* Define $\|A\|^2 := \det A$ for $A \in \mathbb{H}^0$. Compute: \begin{align*} \det A = (i\beta)(-i\beta) - (\gamma + i\delta)(-\gamma + i\delta) = \beta^2 + (\gamma^2 - i^2 \delta^2) = \beta^2 + \gamma^2 + \delta^2. \end{align*} So $\|A\|^2 = \beta^2 + \gamma^2 + \delta^2$ is the standard Euclidean squared norm in coordinates $(\beta, \gamma, \delta)$. Hence $(\mathbb{H}^0, \|\cdot\|)$ is a three-dimensional Euclidean space, naturally identified with $\mathbb{R}^3$ via the basis $(i, j, k)$. We identify $\mathbb{H}^0 \cong \mathbb{R}^3$ via \begin{align*} \Psi: \mathbb{R}^3 &\to \mathbb{H}^0 \\ (\beta, \gamma, \delta) &\mapsto \beta i + \gamma j + \delta k. \end{align*} $\Psi$ is a $\mathbb{R}$-linear bijection that intertwines the Euclidean norm on $\mathbb{R}^3$ with $\|\cdot\|$ on $\mathbb{H}^0$. [/step] [step:Construct the conjugation homomorphism $\varphi: \operatorname{SU}(2) \to \operatorname{O}(3)$] [claim:The map $\Phi: \operatorname{SU}(2) \times \mathbb{H}^0 \to \mathbb{H}^0$, $(X, A) \mapsto X A X^{-1}$, is a well-defined continuous left action by linear isometries] [proof] *Conjugation preserves $\mathbb{H}^0$.* For $X \in \operatorname{SU}(2)$ and $A \in \mathbb{H}^0$, \begin{align*} \operatorname{tr}(X A X^{-1}) = \operatorname{tr}(A X^{-1} X) = \operatorname{tr}(A) = 0, \end{align*} by cyclicity of the trace. So $X A X^{-1} \in \mathbb{H}^0$. *Conjugation preserves $\mathbb{H}$.* The space $\mathbb{H}$ is also characterised as $\{A \in \operatorname{Mat}_2(\mathbb{C}) : A^* = -A + \operatorname{tr}(A) \cdot I\}$, equivalently those matrices for which $A - \frac{1}{2}\operatorname{tr}(A) I$ is anti-Hermitian. Conjugation by $X \in \operatorname{SU}(2)$ (with $X^{-1} = X^*$) preserves the anti-Hermitian-modulo-trace condition: $(X A X^*)^* = X A^* X^*$ and $\operatorname{tr}(XAX^*) = \operatorname{tr}(A)$. Hence conjugation maps $\mathbb{H}$ to $\mathbb{H}$. (This is not strictly needed for the proof but explains why $\mathbb{H}$ is invariant.) *Conjugation is linear.* For $A, B \in \mathbb{H}^0$ and $\lambda \in \mathbb{R}$, $X(\lambda A + B) X^{-1} = \lambda X A X^{-1} + X B X^{-1}$, by distributivity of matrix multiplication. *Conjugation preserves the norm.* For $A \in \mathbb{H}^0$, \begin{align*} \|X A X^{-1}\|^2 = \det(X A X^{-1}) = \det X \cdot \det A \cdot \det X^{-1} = (\det X)(\det X)^{-1} \det A = \det A = \|A\|^2. \end{align*} (Using $\det X = 1$ since $X \in \operatorname{SU}(2)$, plus multiplicativity of determinant.) *Continuity.* The map $(X, A) \mapsto X A X^{-1}$ is a polynomial in the entries of $X, A$ and the entries of $X^{-1}$. The entries of $X^{-1} = X^*$ are continuous functions of the entries of $X$. So $\Phi$ is continuous in $(X, A)$. *Action axioms.* $\Phi(I, A) = A$ and $\Phi(X Y, A) = (XY) A (XY)^{-1} = X (Y A Y^{-1}) X^{-1} = \Phi(X, \Phi(Y, A))$. [/proof] [/claim] By the claim, the map \begin{align*} \varphi: \operatorname{SU}(2) &\to \operatorname{GL}(\mathbb{H}^0) \\ X &\mapsto (A \mapsto X A X^{-1}) \end{align*} is a continuous group homomorphism, and $\varphi(X) \in \operatorname{O}(\mathbb{H}^0, \|\cdot\|)$ for every $X$. Identifying $\mathbb{H}^0 \cong \mathbb{R}^3$ via $\Psi$ and $\operatorname{O}(\mathbb{H}^0, \|\cdot\|) \cong \operatorname{O}(3)$ via the same identification, we obtain a continuous group homomorphism \begin{align*} \varphi: \operatorname{SU}(2) \to \operatorname{O}(3). \end{align*} [guided] The conjugation action $A \mapsto X A X^{-1}$ on $\mathbb{H}^0$ is the *adjoint representation* of $\operatorname{SU}(2)$ on its Lie algebra $\mathfrak{su}(2)$ — the imaginary quaternions $\mathbb{H}^0$ are exactly the trace-zero anti-Hermitian $2 \times 2$ complex matrices, which form $\mathfrak{su}(2)$. So this construction is part of a much wider story: the adjoint representation of any compact Lie group $G$ acts on its Lie algebra $\mathfrak{g}$ by orthogonal transformations under the Killing form. The norm-preservation step is the key calculation. Why does conjugation preserve $\det A$? Because $\det$ is multiplicative: $\det(XAX^{-1}) = \det X \cdot \det A \cdot \det X^{-1} = \det A$, regardless of whether $X$ is unitary. The unitarity is needed for $\mathbb{H}^0$-invariance via the trace argument, but the norm preservation uses only $\det X = 1$. The continuity of $\varphi$ uses that matrix multiplication and matrix inversion are continuous on $\operatorname{GL}_n(\mathbb{C})$ (in particular, on the closed subset $\operatorname{SU}(2)$), so the formula $(X, A) \mapsto X A X^{-1}$ defines a continuous map. [/guided] [/step] [step:Compute the kernel and image: $\ker \varphi = \{\pm I\}$ and $\operatorname{im}\, \varphi \subseteq \operatorname{SO}(3)$] *Kernel.* $X \in \ker \varphi$ iff $X A = A X$ for every $A \in \mathbb{H}^0$, iff $X$ commutes with $i, j, k$. Since $i, j, k$ together generate $\operatorname{Mat}_2(\mathbb{C})$ as a $\mathbb{C}$-algebra (the basis $\{I, i, j, k\}$ spans $\operatorname{Mat}_2(\mathbb{C})$ over $\mathbb{C}$ — a direct check), $X$ commutes with everything in $\operatorname{Mat}_2(\mathbb{C})$ iff $X$ is a scalar matrix. Combined with $X \in \operatorname{SU}(2)$, we have $X = \lambda I$ with $\lambda^2 = \det X = 1$, so $\lambda = \pm 1$. Hence $\ker \varphi = \{\pm I\} = Z(\operatorname{SU}(2))$. *Image lies in $\operatorname{SO}(3)$.* The map $\det \circ \varphi: \operatorname{SU}(2) \to \{\pm 1\}$ is continuous (composition of two continuous maps). Since $\operatorname{SU}(2)$ is path-connected (it is homeomorphic to $S^3$ via $\begin{pmatrix} a & -\bar{b} \\ b & \bar{a} \end{pmatrix} \mapsto (\operatorname{Re} a, \operatorname{Im} a, \operatorname{Re} b, \operatorname{Im} b)$ with $|a|^2 + |b|^2 = 1$, and $S^3$ is path-connected), its continuous image in $\{\pm 1\}$ (a discrete two-point space) must be a single point. Evaluating at $X = I$: $\det(\varphi(I)) = \det(\operatorname{Id}_{\mathbb{H}^0}) = 1$. So $\det(\varphi(X)) = 1$ for every $X \in \operatorname{SU}(2)$, i.e., $\operatorname{im}\, \varphi \subseteq \operatorname{SO}(3)$. [/step] [step:Compute $\varphi$ on three one-parameter subgroups and exhibit rotations in coordinate planes] We compute $\varphi(X)$ for three families of $X$: *Family 1: $X = t_\theta := \mathrm{diag}(e^{i\theta}, e^{-i\theta})$.* For $A = \beta i + \gamma j + \delta k$, \begin{align*} t_\theta i t_\theta^{-1} &= \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \begin{pmatrix} e^{-i\theta} & 0 \\ 0 & e^{i\theta} \end{pmatrix} = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} = i, \\ t_\theta j t_\theta^{-1} &= \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} e^{-i\theta} & 0 \\ 0 & e^{i\theta} \end{pmatrix} = \begin{pmatrix} 0 & e^{2i\theta} \\ -e^{-2i\theta} & 0 \end{pmatrix} \\ &= \cos(2\theta) j + \sin(2\theta) k, \end{align*} where the last step uses $e^{2i\theta} = \cos(2\theta) + i\sin(2\theta)$ and matches against $j = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ and $k = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}$: \begin{align*} \cos(2\theta)j + \sin(2\theta)k = \begin{pmatrix} 0 & \cos(2\theta) + i\sin(2\theta) \\ -\cos(2\theta) + i\sin(2\theta) & 0 \end{pmatrix} = \begin{pmatrix} 0 & e^{2i\theta} \\ -e^{-2i\theta} & 0 \end{pmatrix}. \,\checkmark \end{align*} By a similar computation, $t_\theta k t_\theta^{-1} = -\sin(2\theta) j + \cos(2\theta) k$. So in the basis $(i, j, k)$, the matrix of $\varphi(t_\theta)$ is \begin{align*} \varphi(t_\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(2\theta) & -\sin(2\theta) \\ 0 & \sin(2\theta) & \cos(2\theta) \end{pmatrix}, \end{align*} which is a rotation by angle $2\theta$ in the $(j, k)$-plane (equivalently, rotation about the $i$-axis). *Family 2: $X = r_\theta := \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$.* Verify $r_\theta \in \operatorname{SU}(2)$: $\det r_\theta = \cos^2\theta + \sin^2\theta = 1$, $r_\theta^* r_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} = I$. By a calculation analogous to Family 1 (the structure is identical, with $r_\theta$ acting as a rotation in a different coordinate plane), $\varphi(r_\theta)$ is a rotation by $2\theta$ in the $(i, k)$-plane, i.e., \begin{align*} \varphi(r_\theta) = \begin{pmatrix} \cos(2\theta) & 0 & -\sin(2\theta) \\ 0 & 1 & 0 \\ \sin(2\theta) & 0 & \cos(2\theta) \end{pmatrix}. \end{align*} *Family 3: $X = s_\theta := \begin{pmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{pmatrix}$.* Verify $s_\theta \in \operatorname{SU}(2)$: $\det s_\theta = \cos^2 \theta - i^2 \sin^2\theta = \cos^2\theta + \sin^2\theta = 1$, and $s_\theta^* s_\theta = I$ by direct computation. By an analogous calculation, $\varphi(s_\theta)$ is a rotation by $2\theta$ in the $(i, j)$-plane. [claim:The three families $\{\varphi(t_\theta), \varphi(r_\theta), \varphi(s_\theta) : \theta \in \mathbb{R}\}$ together generate $\operatorname{SO}(3)$ as a group] [proof] Each family produces all rotations in one of the three coordinate planes $(j, k)$, $(i, k)$, $(i, j)$ of $\mathbb{R}^3 \cong \mathbb{H}^0$ (the angles $2\theta$ range over all of $\mathbb{R}/2\pi\mathbb{Z}$ as $\theta$ ranges over $\mathbb{R}/2\pi\mathbb{Z}$, hence they realise every rotation in the coordinate plane). It is a classical fact that rotations in coordinate planes generate $\operatorname{SO}(3)$ (see, e.g., the [Euler-angle decomposition](/theorems/???): every rotation of $\mathbb{R}^3$ is a product of three rotations about coordinate axes, $R = R_z(\alpha) R_y(\beta) R_z(\gamma)$ for some angles $\alpha, \beta, \gamma$). Hence the three families generate $\operatorname{SO}(3)$. [/proof] [/claim] Combining: $\operatorname{im}\, \varphi \supseteq \langle \varphi(t_\theta), \varphi(r_\theta), \varphi(s_\theta) \rangle = \operatorname{SO}(3)$, and by Step 3 $\operatorname{im}\, \varphi \subseteq \operatorname{SO}(3)$, so $\operatorname{im}\, \varphi = \operatorname{SO}(3)$. [guided] The three families $t_\theta, r_\theta, s_\theta$ are the three "natural" one-parameter subgroups of $\operatorname{SU}(2)$ corresponding to the three quaternionic units. Concretely, $t_\theta = \exp(\theta i)$, $r_\theta = \exp(\theta j)$, $s_\theta = \exp(\theta k)$. The fact that conjugation by $\exp(\theta u)$ for a unit imaginary quaternion $u$ acts as rotation by $2\theta$ about the $u$-axis is the *Euler-Rodrigues formula* in disguise. Why is the angle $2\theta$ rather than $\theta$? Because conjugation acts on $\mathbb{H}^0$ as $\exp(2\theta \cdot \mathrm{ad}(u))$ where $\mathrm{ad}(u) = [u, \cdot]$ — the factor of $2$ comes from the bracket $[u, v]_{\mathfrak{su}(2)} = 2(u \times v)$ in the standard normalisation. This "$2$-to-$1$" speed-up is the geometric reason for the kernel being $\{\pm I\}$ rather than just $\{I\}$: the map $\varphi$ winds $\operatorname{SU}(2) \cong S^3$ twice around $\operatorname{SO}(3) \cong \mathbb{RP}^3$. The cited Euler-angle decomposition is a basic fact about $\operatorname{SO}(3)$: every rotation is a product of three rotations about coordinate axes. This is what allows us to claim that rotations in the three coordinate planes generate $\operatorname{SO}(3)$. [/guided] [/step] [step:Pass to the quotient and verify the isomorphism] By the [First Isomorphism Theorem](/theorems/???) for groups, $\varphi$ factors through the quotient $\operatorname{SU}(2) / \ker \varphi = \operatorname{SU}(2) / \{\pm I\}$ as \begin{align*} \bar{\varphi}: \operatorname{SU}(2) / \{\pm I\} &\to \operatorname{SO}(3) \\ [X] &\mapsto X A X^{-1} \text{ (well-defined: independent of choice of representative)}, \end{align*} and $\bar{\varphi}$ is a group isomorphism (algebraically). We verify that $\bar{\varphi}$ is a topological isomorphism (homeomorphism). *$\bar{\varphi}$ is continuous.* The quotient map $\pi: \operatorname{SU}(2) \to \operatorname{SU}(2)/\{\pm I\}$ is continuous (and open) by definition of the quotient topology, and $\varphi = \bar{\varphi} \circ \pi$. So $\bar{\varphi} \circ \pi$ is continuous, which by the universal property of the quotient topology forces $\bar{\varphi}$ continuous. *$\bar{\varphi}$ is a continuous bijection from a compact space to a Hausdorff space.* $\operatorname{SU}(2)$ is compact (it is a closed bounded subset of $\operatorname{Mat}_2(\mathbb{C}) \cong \mathbb{R}^8$), and the quotient $\operatorname{SU}(2)/\{\pm I\}$ is compact as the continuous image of $\operatorname{SU}(2)$ under $\pi$. The codomain $\operatorname{SO}(3)$ is Hausdorff (as a subspace of $\operatorname{Mat}_3(\mathbb{R})$, which is Hausdorff). *$\bar{\varphi}$ is a homeomorphism.* By the closed-map lemma: a continuous bijection $f: X \to Y$ from a compact space $X$ to a Hausdorff space $Y$ is a homeomorphism. (Proof: every closed subset $C \subseteq X$ is compact, $f(C)$ is compact in $Y$ since $f$ is continuous, and compact subsets of Hausdorff spaces are closed. So $f$ is a closed map, and a continuous closed bijection is a homeomorphism.) *$\bar{\varphi}$ is a topological group isomorphism.* It is a homomorphism of abstract groups by Step 4, a homeomorphism by the previous bullet, and these together imply the inverse is also a continuous group homomorphism. Hence $\operatorname{SU}(2)/\{\pm I\}$ and $\operatorname{SO}(3)$ are isomorphic as topological groups. [guided] The compact-to-Hausdorff lemma is the workhorse of algebraic topology in finite-dimensional contexts: it converts an algebraic isomorphism that is continuous in one direction into a topological isomorphism, sparing us from explicitly constructing the inverse and proving its continuity. The lemma fails for non-compact $X$: e.g., the bijection $[0, 2\pi) \to S^1, \theta \mapsto e^{i\theta}$ is continuous but not a homeomorphism, because $[0, 2\pi)$ is not compact. Why is $\operatorname{SO}(3)$ Hausdorff? It is a subspace of $\operatorname{Mat}_3(\mathbb{R})$, which inherits the Euclidean topology from $\mathbb{R}^9$, a Hausdorff space. Subspaces of Hausdorff spaces are Hausdorff. The conclusion of the proof is the natural equivalence \begin{align*} \operatorname{SU}(2)/\{\pm I\} \cong \operatorname{SO}(3), \end{align*} a topological group isomorphism. This identifies $\operatorname{SU}(2)$ as the *spin double cover* of $\operatorname{SO}(3)$: the projection $\varphi: \operatorname{SU}(2) \to \operatorname{SO}(3)$ is a $2$-fold connected covering map (the kernel is $\{\pm I\}$, a discrete two-point group). This double cover is the foundational construction underlying spinors in physics, the Pauli matrices, and the entire theory of spin in quantum mechanics. The "sketch" element of this proof lies in citing two external generation results: (i) that $\{I, i, j, k\}$ spans $\operatorname{Mat}_2(\mathbb{C})$ as a complex algebra (used in the kernel computation), and (ii) that rotations in coordinate planes generate $\operatorname{SO}(3)$ (used in the surjectivity step). Both facts have well-known elementary proofs that can be supplied if needed; the structure of the rest of the argument is fully rigorous. [/guided] [/step]