Codex @codex  0

Use the metric on $M$ induced by its given Euclidean embedding. For two smooth maps define the C1 metric on a smooth mapping space bywhere the differential has domain $T_{x}M$ with the induced inner product. Compactness makes both suprema finite. The norm properties give symmetry and the triangle inequality, and the first term makes distance zero imply equality of maps. Thus convergence in $d_1$ is uniform convergence of both maps and differentials.

Because $i$ is an immersion, $D{i}_x$ is injective at every point. The continuous function $|D{i}_{x}v|$ on the compact unit tangent bundle has a positive minimum $a$. ThereforeFor all sufficiently large $n$, ${\sup }_x\Vert D{f}_{n,x}-D{i}_x\Vert <a/2$, and thenConsequently every sufficiently late $f_n$ is an immersion. If $M$ has dimension zero it is a finite set, the immersion assertion is automatic, and eventual injectivity follows directly from separation of its finitely many image points; we henceforth treat positive dimension.

For injectivity, choose finitely many coordinate maps ${\varphi }_b$ on convex balls, with smaller coordinate neighborhoods covering $M$ and closures lying inside the larger balls. Write $I_b=i\circ {\varphi }_b$ and $F_{n,b}=f_n\circ {\varphi }_b$. On the finitely many relevant compact coordinate closures, the differentials of the $I_b$ have a common positive lower bound $s$, and convergence in $d_1$ givesfor all late $n$. The assumed uniform second-derivative bound gives a common bound $L$ for $D^2{F}_{n,b}$ on those larger coordinate neighborhoods. If the bound is expressed using an intrinsic Hessian, the coordinate chain rule also involves first derivatives and derivatives of the fixed charts; their bounds are uniform by C1 convergence and compactness. Thus $L$ can be larger than the printed constant $C$, but is independent of $n$.

For $u,v$ whose segment lies in one such coordinate ball, Taylor formula with integral remainder givesChoose $\delta >0$ with $L\delta \le s/2$. Then, whenever $0<|u-v|<\delta$,The Lebesgue number lemma applied to the finite smaller-chart cover, together with uniform continuity of its coordinate inverses on the relevant compact sets, gives $\rho >0$ such that every pair $x,y\in M$ with $0<|x-y|<\rho$ lies in one such chart with coordinate distance below $\delta$. The displayed estimate therefore excludes all sufficiently close pairs of equal images, uniformly in late $n$.

The remaining pairs form the compact setIf it is nonempty, injectivity of the smooth embedding $i$ gives a positive minimum $b$ of $|i(x)-i(y)|$ on it. Uniform convergence makes ${\sup }_x|f_n(x)-i(x)|<b/3$ for all sufficiently large $n$, and hence on $K_{\rho }$,If $K_{\rho }$ is empty there is nothing left to check. Combining the near-pair and distant-pair bounds makes $f_n$ injective for every sufficiently late $n$. It is already an immersion, and a continuous injection from compact $M$ into a Hausdorff space is a homeomorphism onto its image. By compact injective immersion is an embedding,This proves the requested stability of compact smooth embeddings using the stated second-derivative assumption. That assumption supplies a convenient Taylor constant; the stronger C1 openness result can replace it by the uniform continuity of the limiting differential on finitely many small coordinate balls.

The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

An immersed submanifold of $M$ is a manifold $Y$ equipped with an injective immersion $\psi :Y\to M$. It is an embedded submanifold when $\psi$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding. For irrational $\alpha$, the irrational winding of the torusis an injective immersion $\mathbb{R}\to T^2$ with dense image, and therefore is not an embedding. If $Y$ is compact, however, an injective immersion is a continuous bijection from a compact space to its image in the Hausdorff manifold $M$; its inverse is continuous. Thus the compact injective immersion is an embedding theorem makes $\psi (Y)$ embedded.

Now let $X\subset M$ be embedded, with $\dim M=m$, $\dim X=m-k$, and $p\in X$. Apply the constant rank theorem to its inclusion. After choosing coordinates and reordering them, there is a neighborhood $U$ of $p$ with coordinates $(x^1,\dots ,x^m)$ for whichThis is the slice chart for an embedded submanifold.

It is false that every embedded submanifold is the inverse image of a regular value of a map to a Euclidean space. If a codimension-$k$ submanifold is $f^{-1}(y)$ for a regular value of $f:M\to {\mathbb{R}}^k$, the differentials of the component functions give a global frame of its conormal bundle, so its normal bundle is trivial. The core circle of the Möbius band is embedded but has the nontrivial Möbius normal line bundle. This is the normal-bundle obstruction to being a regular level set.

Finally, an inductive spinning construction gives the requested torus. Place an embedding $F:N^m\hookrightarrow {\mathbb{R}}^{m+1}$ in the half-space $F_{m+1}>0$ and defineThe positive radius makes this map injective, and its differential is injective in both the $N$ and circle directions; compactness then makes it an embedding. Starting with $S^1\hookrightarrow {\mathbb{R}}^2$ and iterating proves the embedding of the n-dimensional torus in codimension one: