[proofplan]
We construct the homomorphism by letting $SU(2)$ act by conjugation on the real three-dimensional [vector space](/page/Vector%20Space) of traceless Hermitian $2\times 2$ complex matrices. The trace [inner product](/page/Inner%20Product) identifies this space with Euclidean $\mathbb R^3$, and conjugation preserves the inner product and orientation, so the action gives a map into $SO(3)$. We compute the kernel by showing that a matrix commuting with all traceless Hermitian matrices must be scalar, and then use explicit exponentials to show that the image contains the standard rotations generating $SO(3)$. The quotient statement follows from the universal property of the quotient and the smooth bijective homomorphism induced by $\Phi$.
[/proofplan]
[step:Identify $\mathbb R^3$ with traceless Hermitian matrices]
Let $V$ denote the real vector space
\begin{align*}
V=\{X\in M(2,\mathbb C):X^*=X,\operatorname{tr}X=0\}.
\end{align*}
Define matrices
\begin{align*}
E_1=\begin{pmatrix}1&0
\end{align*}
\begin{align*}
0&-1\end{pmatrix},\qquad
E_2=\begin{pmatrix}0&1
\end{align*}
\begin{align*}
1&0\end{pmatrix},\qquad
E_3=\begin{pmatrix}0&i
\end{align*}
\begin{align*}
-i&0\end{pmatrix}.
\end{align*}
Each $E_j$ lies in $V$, and every $X\in V$ has a unique expression
\begin{align*}
X=x_1E_1+x_2E_2+x_3E_3
\end{align*}
with $x_1,x_2,x_3\in\mathbb R$. Thus the map
\begin{align*}
\theta:\mathbb R^3&\to V
\end{align*}
\begin{align*}
(x_1,x_2,x_3)&\mapsto x_1E_1+x_2E_2+x_3E_3
\end{align*}
is a real vector-space isomorphism.
Define an inner product on $V$ by
\begin{align*}
\langle X,Y\rangle_V=\frac{1}{2}\operatorname{tr}(XY)
\end{align*}
for $X,Y\in V$. Since $E_jE_j=I_2$ and $\operatorname{tr}(E_jE_k)=0$ for $j\neq k$, the ordered basis $(E_1,E_2,E_3)$ is orthonormal for $\langle\cdot,\cdot\rangle_V$. Therefore $\theta$ is an orientation-preserving isometry from Euclidean $\mathbb R^3$ onto $V$.
[/step]
[step:Define the conjugation homomorphism and show that it lands in $SO(3)$]
For $U\in SU(2)$, define a real-[linear map](/page/Linear%20Map)
\begin{align*}
\Phi(U):V&\to V
\end{align*}
\begin{align*}
X&\mapsto UXU^{-1}.
\end{align*}
This is well-defined because $U^{-1}=U^*$ for $U\in SU(2)$. If $X=X^*$, then
\begin{align*}
(UXU^{-1})^*=UXU^{-1},
\end{align*}
and
\begin{align*}
\operatorname{tr}(UXU^{-1})=\operatorname{tr}(XU^{-1}U)=\operatorname{tr}X=0
\end{align*}
by cyclicity of trace. Hence $\Phi(U)(X)\in V$.
For $X,Y\in V$, cyclicity of trace gives
\begin{align*}
\langle \Phi(U)X,\Phi(U)Y\rangle_V
=\frac{1}{2}\operatorname{tr}(UXU^{-1}UYU^{-1})
=\frac{1}{2}\operatorname{tr}(UXYU^{-1})
=\frac{1}{2}\operatorname{tr}(XY)
=\langle X,Y\rangle_V.
\end{align*}
Thus $\Phi(U)\in O(V)$.
The map $U\mapsto \det(\Phi(U))$ is continuous from $SU(2)$ to $\{\pm1\}$. The group $SU(2)$ is connected: every element of $SU(2)$ has the form
\begin{align*}
\begin{pmatrix}\alpha&\beta
\end{align*}
\begin{align*}
-\overline{\beta}&\overline{\alpha}\end{pmatrix}
\end{align*}
with $\alpha,\beta\in\mathbb C$ and $|\alpha|^2+|\beta|^2=1$, so $SU(2)$ is homeomorphic to the unit sphere $S^3\subset\mathbb R^4$. Since $\Phi(I_2)=\operatorname{id}_V$, we have $\det(\Phi(I_2))=1$, and connectedness forces $\det(\Phi(U))=1$ for all $U\in SU(2)$. Therefore $\Phi$ lands in $SO(V)$, which is identified with $SO(3)$ by the oriented [orthonormal basis](/page/Orthonormal%20Basis) $(E_1,E_2,E_3)$.
Finally, $\Phi$ is a [group homomorphism](/page/Group%20Homomorphism) because for $U,W\in SU(2)$ and $X\in V$,
\begin{align*}
\Phi(UW)(X)=UW X (UW)^{-1}=U(WXW^{-1})U^{-1}=\Phi(U)(\Phi(W)(X)).
\end{align*}
Its matrix entries in the basis $(E_1,E_2,E_3)$ are polynomial expressions in the real and imaginary parts of the entries of $U$, so $\Phi:SU(2)\to SO(3)$ is smooth.
[guided]
The idea is to turn a two-dimensional complex unitary matrix into a three-dimensional real rotation by asking how it conjugates traceless Hermitian matrices. For $U\in SU(2)$, we define
\begin{align*}
\Phi(U):V&\to V
\end{align*}
\begin{align*}
X&\mapsto UXU^{-1}.
\end{align*}
We must first check that this formula actually maps $V$ into itself. Since $U\in SU(2)$ is unitary, $U^{-1}=U^*$. If $X=X^*$, then
\begin{align*}
(UXU^{-1})^*=(U^{-1})^*X^*U^*=UXU^{-1}.
\end{align*}
Thus $UXU^{-1}$ is Hermitian. Also, cyclicity of trace gives
\begin{align*}
\operatorname{tr}(UXU^{-1})=\operatorname{tr}(XU^{-1}U)=\operatorname{tr}X=0.
\end{align*}
So $UXU^{-1}\in V$.
Next we check that this real-linear map is an isometry. For $X,Y\in V$, the trace inner product satisfies
\begin{align*}
\langle \Phi(U)X,\Phi(U)Y\rangle_V
=\frac{1}{2}\operatorname{tr}(UXU^{-1}UYU^{-1})
=\frac{1}{2}\operatorname{tr}(UXYU^{-1})
=\frac{1}{2}\operatorname{tr}(XY)
=\langle X,Y\rangle_V.
\end{align*}
The cancellation $U^{-1}U=I_2$ occurs in the middle, and the final equality again uses cyclicity of trace. Therefore $\Phi(U)$ is orthogonal.
An orthogonal transformation may have determinant $1$ or $-1$, so we still need orientation preservation. The determinant map
\begin{align*}
U\mapsto \det(\Phi(U))
\end{align*}
is continuous and takes values in the discrete set $\{\pm1\}$. The group $SU(2)$ is connected because every element has the form
\begin{align*}
\begin{pmatrix}\alpha&\beta
\end{align*}
\begin{align*}
-\overline{\beta}&\overline{\alpha}\end{pmatrix}
\end{align*}
with $|\alpha|^2+|\beta|^2=1$, identifying $SU(2)$ with the connected sphere $S^3$. Since $\Phi(I_2)=\operatorname{id}_V$ has determinant $1$, the determinant cannot jump to $-1$. Hence $\Phi(U)\in SO(V)$ for every $U\in SU(2)$.
The homomorphism identity is a direct computation:
\begin{align*}
\Phi(UW)(X)=UW X (UW)^{-1}=U(WXW^{-1})U^{-1}=\Phi(U)(\Phi(W)(X)).
\end{align*}
Smoothness follows because, after writing $\Phi(U)(E_j)$ in the basis $(E_1,E_2,E_3)$, the coefficients are polynomial expressions in the real and imaginary parts of the entries of $U$.
[/guided]
[/step]
[step:Compute the kernel as the two central scalar matrices]
Suppose $U\in\ker\Phi$. Then $UXU^{-1}=X$ for every $X\in V$, equivalently $UX=XU$ for every $X\in V$. In particular $U$ commutes with $E_1$ and $E_2$.
Write
\begin{align*}
U=\begin{pmatrix}a&b
\end{align*}
\begin{align*}
c&d\end{pmatrix}
\end{align*}
with $a,b,c,d\in\mathbb C$. The identity $UE_1=E_1U$ gives
\begin{align*}
\begin{pmatrix}a&-b
\end{align*}
\begin{align*}
c&-d\end{pmatrix}
=
\begin{pmatrix}a&b
\end{align*}
\begin{align*}
-c&-d\end{pmatrix},
\end{align*}
so $b=0$ and $c=0$. Thus $U=\operatorname{diag}(a,d)$. The identity $UE_2=E_2U$ then gives
\begin{align*}
\begin{pmatrix}0&a
\end{align*}
\begin{align*}
d&0\end{pmatrix}
=
\begin{pmatrix}0&d
\end{align*}
\begin{align*}
a&0\end{pmatrix},
\end{align*}
so $a=d$. Hence $U=aI_2$. Since $U\in SU(2)$, $\det U=a^2=1$, and therefore $a=\pm1$. Thus
\begin{align*}
\ker\Phi=\{I_2,-I_2\}.
\end{align*}
The reverse inclusion holds because both $I_2$ and $-I_2$ commute with every $X\in V$.
[/step]
[step:Show that the image contains the standard coordinate rotations]
For $j\in\{1,2,3\}$ and $t\in\mathbb R$, define
\begin{align*}
U_j(t)=\cos(t/2)I_2+i\sin(t/2)E_j.
\end{align*}
Since $E_j^*=E_j$ and $E_j^2=I_2$, we have
\begin{align*}
U_j(t)^*U_j(t)=I_2
\end{align*}
and
\begin{align*}
\det U_j(t)=1,
\end{align*}
so $U_j(t)\in SU(2)$.
Using the multiplication relations
\begin{align*}
E_1E_2=-iE_3,\qquad E_2E_3=-iE_1,\qquad E_3E_1=-iE_2
\end{align*}
and the reversed relations with sign changed, direct multiplication gives
\begin{align*}
\Phi(U_1(t))(E_1)=E_1,
\end{align*}
\begin{align*}
\Phi(U_1(t))(E_2)=\cos t\,E_2+\sin t\,E_3,
\end{align*}
and
\begin{align*}
\Phi(U_1(t))(E_3)=-\sin t\,E_2+\cos t\,E_3.
\end{align*}
Thus $\Phi(U_1(t))$ is the rotation about the $E_1$-axis through angle $t$ in the oriented orthonormal basis $(E_1,E_2,E_3)$. The same computation with cyclically permuted indices shows that the image of $\Phi$ contains all rotations about each of the three coordinate axes.
[/step]
[step:Generate every element of $SO(3)$ from coordinate-axis rotations]
Let $e_1,e_2,e_3\in\mathbb R^3$ denote the standard ordered basis, and let $R_j(s)\in SO(3)$ denote the rotation of $\mathbb R^3$ about the axis $\mathbb R e_j$ through angle $s$. Let $R\in SO(3)$. If $Re_3\neq -e_3$, choose $\alpha\in\mathbb R$ and $\beta\in\mathbb R$ such that
\begin{align*}
R_3(\alpha)R_2(\beta)e_3=Re_3.
\end{align*}
Then
\begin{align*}
R_2(-\beta)R_3(-\alpha)R
\end{align*}
fixes $e_3$ and lies in $SO(3)$, hence it is a rotation $R_3(\gamma)$ about the third coordinate axis for some $\gamma\in\mathbb R$. Therefore
\begin{align*}
R=R_3(\alpha)R_2(\beta)R_3(\gamma).
\end{align*}
If $Re_3=-e_3$, set $\beta=\pi$. Then $R_2(\pi)e_3=-e_3$, so $R_2(-\pi)R$ fixes $e_3$. Its restriction to the oriented plane $e_3^\perp$ is an orientation-preserving orthogonal map of a two-dimensional Euclidean space, hence is a rotation by some angle $\gamma\in\mathbb R$. Therefore $R_2(-\pi)R=R_3(\gamma)$, and hence
\begin{align*}
R=R_2(\pi)R_3(\gamma).
\end{align*}
This is again a product of coordinate-axis rotations.
By the preceding step, each coordinate-axis rotation lies in $\operatorname{im}\Phi$. Since $\operatorname{im}\Phi$ is a subgroup of $SO(3)$, every product of coordinate-axis rotations lies in $\operatorname{im}\Phi$. Hence $\Phi$ is surjective.
[/step]
[step:Pass to the quotient by the kernel]
Since $\ker\Phi=\{I_2,-I_2\}$, the map $\Phi$ is constant on left cosets of $\{I_2,-I_2\}$. Therefore it induces a smooth homomorphism
\begin{align*}
\overline{\Phi}:SU(2)/\{I_2,-I_2\}&\to SO(3)
\end{align*}
\begin{align*}
U\{I_2,-I_2\}&\mapsto \Phi(U).
\end{align*}
This induced map is well-defined precisely because multiplying $U$ by an element of the kernel does not change $\Phi(U)$. It is surjective because $\Phi$ is surjective, and it is injective because, with $I_3$ denoting the $3\times3$ identity matrix,
\begin{align*}
\overline{\Phi}(U\{I_2,-I_2\})=I_3
\end{align*}
implies $U\in\ker\Phi=\{I_2,-I_2\}$, so the coset is the identity coset.
The subgroup $\{I_2,-I_2\}$ is closed and normal in $SU(2)$ because it is the kernel of the continuous homomorphism $\Phi$, so the closed [normal subgroup](/page/Normal%20Subgroup) quotient theorem gives $SU(2)/\{I_2,-I_2\}$ its quotient Lie group structure and makes the projection $q:SU(2)\to SU(2)/\{I_2,-I_2\}$ a smooth quotient homomorphism. The factorisation identity $\Phi=\overline{\Phi}\circ q$ gives smoothness of $\overline{\Phi}$. Since $\overline{\Phi}$ is a bijective smooth homomorphism between finite-dimensional Lie groups, the standard inverse theorem for Lie group homomorphisms implies that $\overline{\Phi}^{-1}$ is smooth. Hence
\begin{align*}
SU(2)/\{I_2,-I_2\}\cong SO(3)
\end{align*}
as Lie groups.
[/step]
