= Solution
The two <Lie groups> have the same local infinitesimal structure, but different global topology. This difference determines which <Lie algebra representations> integrate to representations of each group.
An element of the <SU(2) group> is a <unitary matrix> of <determinant> one. Orthogonality of its columns and its <determinant> give the unique form
$$
U=\begin{pmatrix}a&b\\-\overline b&\overline a\end{pmatrix},\qquad |a|^2+|b|^2=1.
$$
Thus its <group manifold> is the unit <three-sphere> in $\mathbb C^2\simeq\mathbb R^4$: this is <SU(2) as the three-sphere>. In particular it is <compact>, <connected> and <simply connected>. In terms of the <Pauli matrices> one may also write $U=a_0I-i\mathbf a\cdot\boldsymbol\sigma$ with real coefficients satisfying $a_0^2+\mathbf a^2=1$.
Differentiate $U(t)^\dagger U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The <tangent space> consists of traceless <skew-Hermitian matrices>:
$$
\mathfrak{su}(2)=\{X\in M_2(\mathbb C):X^\dagger=-X,\ \operatorname{tr}X=0\}.
$$
The <Lie bracket> of a <Matrix Lie group> is the matrix <commutator>. With $T_a=-i\sigma_a/2$, the <Pauli matrix multiplication law> gives
$$
[T_a,T_b]=\epsilon_{abc}T_c.
$$
This derives the <SU(2) Lie algebra> as a three-dimensional real <Lie algebra>. The Hermitian physics generators $J_a=\sigma_a/2$ instead obey $[J_a,J_b]=i\epsilon_{abc}J_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.
The <SO(3) group> consists of real <orthogonal matrices> with <determinant> one. Differentiating $R(t)^TR(t)=I$ at the identity gives
$$
\mathfrak{so}(3)=\{A\in M_3(\mathbb R):A^T=-A\}.
$$
The <determinant> condition gives no additional infinitesimal constraint because a <skew-symmetric matrix> already has <trace> zero. Define $L_a\mathbf v=\mathbf e_a\times\mathbf v$. The <cross product> identity implies
$$
[L_a,L_b]\mathbf v=\mathbf e_a\times(\mathbf e_b\times\mathbf v)-\mathbf e_b\times(\mathbf e_a\times\mathbf v)=\epsilon_{abc}L_c\mathbf v.
$$
Therefore the <SO(3) Lie algebra> has the same <structure constants> and $T_a\mapsto L_a$ is a <Lie algebra isomorphism>.
The global relation is the <Adjoint double cover from SU(2) to SO(3)>. For $V=\mathbf v\cdot\boldsymbol\sigma$, define $R(U)$ by
$$
UVU^{-1}=(R(U)\mathbf v)\cdot\boldsymbol\sigma.
$$
Conjugation preserves the real space of traceless <Hermitian matrices> and its <inner product> $\tfrac12\operatorname{tr}(VW)=\mathbf v\cdot\mathbf w$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a <group homomorphism>. If $R(U)=I$, then $U$ commutes with every <Pauli matrix>, hence is scalar; unitarity and <determinant> one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,
$$
U(\theta,\mathbf n)=\exp\left(-\frac{i\theta}{2}\mathbf n\cdot\boldsymbol\sigma\right)=\cos\frac\theta2\,I-i\sin\frac\theta2\,\mathbf n\cdot\boldsymbol\sigma
$$
induces rotation through angle $\theta$ about $\mathbf n$, by the <Rodrigues rotation formula>. Every three-dimensional rotation has such an axis and angle, proving surjectivity. Consequently
$$
\boxed{SO(3)\simeq SU(2)/\{\pm I\},\qquad \mathfrak{so}(3)\simeq\mathfrak{su}(2).}
$$
The matrices $U$ and $-U$ are antipodal points on the three-sphere, so the <SO(3) group manifold> is <Real projective space> $\mathbb{RP}^3$. Equivalently, the closed axis-angle ball $\|\theta\mathbf n\|\le\pi$ has opposite boundary points identified. The <fundamental group> is $\pi_1(SO(3))\simeq\mathbb Z_2$, whereas $\pi_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the <universal cover> and the groups are not globally isomorphic.
For <representation theory>, specify finite-dimensional complex continuous representations. Compactness permits an invariant <Hermitian inner product>, obtained by averaging against <Haar measure>, and therefore complete reducibility. Complexifying either real <Lie algebra> gives the <sl2 Lie algebra>. Its finite-dimensional irreducibles are indexed by $n\in\mathbb Z_{\ge0}$, have <highest weight> $n$, and have dimension $n+1$. By <integration of a Lie-algebra representation>, since $SU(2)$ is <simply connected>, every such <Lie algebra representation> integrates uniquely. The resulting <homogeneous polynomial representation of SU2> is
$$
V_n=\operatorname{Sym}^n(\mathbb C^2),\qquad \dim V_n=n+1.
$$
In the <spin angular momentum> notation $j=n/2$, its Hermitian $J_3$ <eigenvalues> are $j,j-1,\ldots,-j$. The central matrix $-I$ acts on the <symmetric power> by $(-1)^n$, so <descent of an SU(2) representation to SO(3)> occurs exactly when $n$ is even. Hence
$$
\boxed{SU(2):j=0,\tfrac12,1,\tfrac32,\ldots;\qquad SO(3):j=0,1,2,\ldots;\qquad \dim V_j=2j+1.}
$$
For a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic <Lie algebras>.
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