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Let $m=\dim M$, $n=\dim N$ and $k={\mathrm{codim}}_{N}S$. Fix $p\in f^{-1}(S)$ and choose the allowed slice chart for an embedded submanifold on a neighborhood $U$ of $f(p)$, so that $S\cap U$ is given by $x^1=\cdots =x^k=0$. On $f^{-1}(U)$ define the smooth mapThe derivative $L=d(x^1,\dots ,x^k)$ at a point of $S\cap U$ is surjective and has kernel $TS$, by the slice description. The printed sum-of-spaces condition givesIndeed any target tangent vector is a sum of an image vector and a tangent vector to $S$, and $L$ kills the latter. By the chain rule, $d{G}_r=L\circ d{f}_r$ is therefore surjective. This condition is precisely transversality of a map to a submanifold; the original sum need not be direct.

We have $G^{-1}(0)=f^{-1}(S)\cap f^{-1}(U)$. Part (b), applied to $G$, gives slice charts for an embedded submanifold of codimension $k$ around every point of this zero set. Since the construction is available around every $p\in f^{-1}(S)$, the charts give its global structure as an embedded submanifold with the subspace topology. Its dimension is $\boxed{m-k=m-n+\dim S}$. Moreover,This follows either from the slice coordinates or from $T_p{G}^{-1}(0)=\ker d{G}_p$ and $\ker L=T_{f(p)}S$. No injectivity of $df$, closedness of $S$, or direct-sum hypothesis is required.

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:

Because the inclusion $\iota :N\hookrightarrow M$ has constant rank $n$, the constant rank theorem gives coordinates $y^1,\dots ,y^n$ on $N$ and $x^1,\dots ,x^m$ on $M$ in whichSince $\iota$ is an embedding, the ambient chart can be shrunk so that it meets no other local sheet of $N$. It then satisfiesand $(x^1,\dots ,x^n)$ restricts to the required chart on $N$. This is a slice chart for an embedded submanifold.

If $X$ is tangent to $N$ and $f{|}_N=0$, then the restriction of $f$ to every curve in $N$ is zero, so $(Xf){|}_N=0$.

Conversely, use a slice chart for an embedded submanifold. The functions $x^{n+1},\dots ,x^m$ vanish on $N$. Writing $X=X^i{\partial }_i$, the hypothesis givesThus $X$ has no normal component and is tangent to $N$. Equivalently, $X$ preserves the vanishing ideal of an embedded submanifold.

A smooth map $f:M\to N$ is transverse to an embedded submanifold $S\subset N$ whenThis is a sum of vector subspaces, not necessarily a direct sum. Equivalently the composition of $d{f}_p$ with the quotient to $T_{f(p)}N/T_{f(p)}S$ is surjective. In a slice chart for an embedded submanifold of codimension $k$, the $k$ normal coordinate functions composed with $f$ have surjective derivative on their common zero set. If the preimage is empty the condition holds vacuously.

If $f:M\to N$ satisfies transversality of a map to a submanifold $S$, its nonempty preimage is an embedded submanifold of the displayed codimension. Choose a slice chart for an embedded submanifold around $f(p)$ in which $S$ is the zero set of the first $k$ coordinates. Their composition with $f$ has surjective derivative on its zero set, so the regular level set theorem gives a local embedded slice of dimension $\dim M-k$. These slices give the subspace topology and a compatible smooth atlas on the preimage. Its tangent space is