Characteristic-class vanishing is only necessary for embedding

Assume AC. Remark. For a closed smooth $m$-manifold every embedding into ${\mathbb{R}}^n$ with $n\ge m$ has a rank-$(n-m)$ normal bundle, so the vanishing of all normal Stiefel-Whitney classes in degrees $i\ge n-m$ (when $m\ge 1$ and $n-m\ge 1$), of the Euler class of an oriented embedded normal bundle (when $m\ge 1$ and $n-m\ge 1$), and of all normal Pontryagin classes with $2i>n-m$ is necessary for embeddability; the top Stiefel–Whitney and oriented Euler vanishings are embedding-specific here (Embedding obstructions include all immersion normal-class obstructions). It is not sufficient: ${\mathbb{R}\mathbb{P}}^3$ has a rank-one trivial stable normal inverse. Indeed, for $q=(q_0,q_1,q_2,q_3)\in S^3$, the three fields $(-q_1,q_0,-q_3,q_2)$, $(-q_2,q_3,q_0,-q_1)$ and $(-q_3,-q_2,q_1,q_0)$ are perpendicular to $q$ and pairwise orthonormal, by direct dot products, so frame $T_q{S}^3$ (The tangent space of a regular level set is the kernel). Each satisfies $X(-q)=-X(q)$, so their differentials descend through $S^3\to {\mathbb{R}\mathbb{P}}^3$. This map is a local diffeomorphism: on an affine chart its two local inverses are the representatives with the chosen coordinate $1$, normalized to unit length with the two signs (Real projective space from affine charts). The descended fields therefore give a smooth tangent frame. Thus the parallelizable-manifold proposition gives $\bar{w}=\bar{p}=1$ and a trivial rank-one inverse; its Euler class is zero by its constant nonzero section (Parallelizable manifolds have no stable characteristic-class obstruction to Euclidean immersion, A nowhere-zero section forces the Euler class to vanish). Nevertheless ${\mathbb{R}\mathbb{P}}^3$ does not smoothly embed in ${\mathbb{R}}^4$: $H_1({\mathbb{R}\mathbb{P}}^3;\mathbb{Z})=\mathbb{Z}\mathrm{/}2$. To compute this, use one cell in each dimension $0,1,2,3$ in the standard quotient cellulation. The 1-cell has coinciding endpoints, so $d_1=0$, and the boundary of the 2-cell maps twice around ${\mathbb{R}\mathbb{P}}^1$, so $d_2=2$ with consistent orientations; hence $\ker d_1\mathrm{/}\mathrm{im}{d}_2=\mathbb{Z}\mathrm{/}2$ by the cellular incidence and homology theorems (Cellular boundary is the incidence degree matrix, Cellular homology computes singular homology; Hatcher, Algebraic Topology, Example 2.42, printed p.144). For the 3-cell, the two hemispheres of its attaching sphere cover the open 2-cell once each. Their signed incidence contributions are $1$ and $-1$: the antipodal identification on $S^2$ has degree $(-1{)}^3=-1$ (Degree of identity constant reflection and antipodal sphere maps). Thus $d_3=1-1=0$ by the cellular incidence theorem, so $H_2({\mathbb{R}\mathbb{P}}^3;\mathbb{Z})=0$ and $H_3({\mathbb{R}\mathbb{P}}^3;\mathbb{Z})=\mathbb{Z}$ by the same cellular homology theorem. The descended tangent frame orients this closed connected three-manifold. The proved local obstruction A closed three-manifold with H_1 = Z/2 and H_2 = 0 does not embed in S^4 therefore rules out a smooth embedding in $S^4$, and hence one in ${\mathbb{R}}^4$. Its proof uses the two complementary sides, Mayer–Vietoris, Alexander duality and UCT; no classification of embeddings is required.

Separately, characteristic classes do not classify embeddings up to isotopy. Let $i:S^2\hookrightarrow {\mathbb{R}}^3$ be the unit sphere inclusion and $r(x_1,x_2,x_3)=(x_1,x_2,-x_3)$. Their radial fields $n_i(x)=x$ and $n_r(x)=r(x)$ trivialize their normal lines: reflection preserves inner products, so $⟨r(x),d{r}_x(v)⟩=⟨x,v⟩=0$ for tangent vectors $v$. Thus the two embeddings have identical stable characteristic classes. If they were isotopic, flattening time near the endpoints and applying The isotopy extension theorem would give an ambient isotopy with time-one diffeomorphism $H$ satisfying $H\circ i=r$. Continuity of the nonzero differential determinant along the isotopy makes $H$ orientation preserving. Since $H(S^2)=S^2$, it permutes the two complementary components; it preserves the bounded ball component because the closure of its image is $H(B^3)$, which is compact, while the exterior has unbounded closure. Therefore $H(B^3)=B^3$. An orientation-preserving diffeomorphism of the ball carries outward-pointing boundary vectors to outward-pointing vectors and preserves its induced boundary orientation (Induced boundary orientation). Its restriction to $S^2$ has degree $1$ (Degree of an orientation-preserving or reversing diffeomorphism), whereas $H{\mathrm{\mid }}_{S^2}=r$ has degree $-1$ (Degree of identity constant reflection and antipodal sphere maps), a contradiction. This proves the failure of isotopy classification by these classes. Isotopy extension applies to an isotopy already given and supplies no criterion for its existence.