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A smooth manifold of dimension $k$ is a Hausdorff, second-countable space with an atlas of smoothly compatible charts to open subsets of ${\mathbb{R}}^k$. A value $y\in Y$ is a regular value of a smooth map $f:X\to Y$ when every $x\in f^{-1}(y)$ has surjective derivative $d{f}_x$.

The inverse function theorem says that a smooth map between equal-dimensional manifolds whose derivative is invertible at a point is a diffeomorphism between neighbourhoods of that point and its image. If $y$ is a regular value of a map from an $m$-manifold to an $n$-manifold, choose coordinates corresponding to an invertible $n$ by $n$ minor of $df$. Applying the inverse theorem to $f$ together with the remaining $m-n$ coordinates makes $f$ the coordinate projection onto the first $n$ coordinates. Its level set is therefore locally ${\mathbb{R}}^{m-n}$. This proves the preimage theorem.

The critical-point set is closed because failure of full rank is the simultaneous vanishing of all maximal minors of $df$. If $X$ is compact, this set is compact, so its image under $f$ is compact and hence closed in the manifold $Y$. Its complement, the set of regular values, is open.

For the printed equations putA rank calculation shows that $(a,0)$ is a regular value when $a>0$: if
$\nabla F_2=\lambda \nabla F_1$, thenAt a point of the level set these relations force $a\le 0$. Hence for $a>0$, the preimage theorem makes $X_a$ a two-dimensional manifold.

As printed, however, the assertion for every $a\ne 0$ is false. If $a<0$, the pointsbelong to $X_a$. Near either point the first equation solves smoothly for $w$, while the second equation isIts positive-$z$ and negative-$z$ sheets meet only at the origin, so deleting the meeting point disconnects every sufficiently small neighbourhood. A punctured neighbourhood in a two-manifold is connected. Thus $X_a$ is not a manifold there.

For $a=0$, Cauchy--Schwarz givesThe first equation requires $w^2=x+y-z^2\ge 0$, so equality must hold. ThereforeThis is the image of a smooth embedding with nonzero derivative, and hence
$\boxed{X_0\text{ is a one-dimensional manifold}}$.

Assume first that $Y$ is connected; otherwise the definition is made separately on each connected component. Choose a regular value $y\in Y$ and define the degree modulo two by

The following course results make this well-defined.

The same transverse-preimage argument applied to a smooth homotopy proves homotopy invariance. Consequently the definition depends only on $f$, rather than on the chosen regular value.

A smooth map $i:N\to M$ is a smooth embedding when it is an injective immersion and a homeomorphism onto its image with the subspace topology. The unit sphere is the inverse image of the regular value $1$ under $x\mapsto |x{|}^2$, so the preimage theorem makes it a smooth submanifold of ${\mathbb{R}}^{n+1}$ and its inclusion an immersion. It is injective, and a continuous injection from the compact sphere into the Hausdorff Euclidean space is a homeomorphism onto its image. Thus the inclusion is an embedding.

Products of spheres can be embedded by iterated spinning. Start with a round $S^{n_1}\subset {\mathbb{R}}^{n_1+1}$ translated into the half-space whose last coordinate $r$ is positive. If a compact $m$-manifold $N$ is embedded bythenis an injective immersion; compactness again makes it an embedding. Iterating with $q=n_2,\dots ,n_k$ embeds $S^{n_1}\times \cdots \times S^{n_k}$ in ${\mathbb{R}}^{n+1}$, where $n={\sum }_i{n}_i$.

Let ${\mathrm{Skew}}_{2n}$ be the skew-symmetric $2n$ by $2n$ matrices and defineThen $S{p}_n(\mathbb{R})=F^{-1}(J)$, andAt a symplectic $A$, put $H=AX$. The derivative becomes $X^{T}J+JX$. Given skew $K$, choosing $X=-\frac{1}{2}JK$ gives $X^{T}J+JX=K$, so the derivative is surjective.

The preimage theorem makes the level set a submanifold. Sinceits dimension is $2n^2+n$.

A value $y$ is regular when $d{f}_x:T_{x}X\to T_{y}Y$ is surjective for every $x\in f^{-1}(y)$. The preimage theorem says $f^{-1}(y)$ is then a submanifold of dimension $\dim X-\dim Y$.

Define $F:TX\to \mathbb{R}$ by $F(p,v)=I_p(v,v)$. On $F^{-1}(1)$, the vertical derivative in direction $v$ is $2I_p(v,v)=2$, so 1 is regular. Since $\dim TX=4$, the preimage theorem makes the unit tangent bundle $SX$ a 3-manifold.