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Imagine folding a sheet along its horizontal axis by the smooth map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$, $F(x,y)=(x,y^2)$. The inputs $(a,r)$ and $(a,-r)$ have the same output. Along $y=0$, horizontal motion still changes the output, but vertical motion produces no first-order change: $F(a,\epsilon )-F(a,0)=(0,{\epsilon }^2)$. These inputs are critical points of a smooth map even though their differential is not the zero linear map. The missing direction in the target tangent space is what matters.

Precisely, let $f:X^m\to Y^n$ be a smooth map between smooth manifolds without boundary. An input $p\in X$ is a critical point of a smooth map when its differentialis not surjective, equivalently when its rank of a linear map is less than $n$. An input where it is surjective is a regular point; a map with no critical points of a smooth map is a submersion. The threshold is the target dimension, not $\min (m,n)$. In manifold charts, compute the matrix rank of the $n\times m$ Jacobian matrix; changes of charts preserve that matrix rank. ^[ref]

For the fold,Thus its set of critical points of a smooth map is the entire horizontal axis. Away from that axis, its differential can produce every infinitesimal target displacement. At the axis, it can produce only horizontal ones. This diagnoses where the two inverse branches in the picture merge.

For a scalar smooth function $f:X\to \mathbb{R}$, the target dimension is one. A linear map to $\mathbb{R}$ fails to be surjective exactly when it is zero, so here a critical point of a smooth map means $d{f}_p=0$. In Euclidean space this is precisely the familiar critical point condition $\nabla f(p)=0$: every first partial derivative vanishes. On a Riemannian manifold, it is equivalently ${\nabla }_{g}f(p)=0$ for the chosen Riemannian metric; the Riemannian gradient depends on that metric, but whether it vanishes does not. See Loring W. Tu, An Introduction to Manifolds, Proposition 8.23 for the scalar coordinate criterion. The scalar smooth function $f(t)=t^3$ has a critical point at zero without a local maximum or local minimum.

A source of lower dimension gives a different warning. For the smooth map $G:\mathbb{R}\to {\mathbb{R}}^2$, $G(t)=(t,t^2)$, the Jacobian matrix is the column $(1,2t{)}^T$, of matrix rank one at every input. Every input is a critical point of a smooth map, since the target dimension is two, even though this parametric curve has nonzero velocity everywhere and is an immersion. More generally, if $m<n$, no differential $T_{p}X\to T_{f(p)}Y$ can be surjective, so every input is critical. The terminology measures the ability to fill target directions, rather than whether the parametrized curve looks smooth.

Every proper surjective submersion is a locally trivial smooth fiber bundle. In particular, the fibers in a proper smooth family are diffeomorphic.

Functions whose differentials are linearly independent at the point in question. For $F=(F_1,\dots ,F_r)$ this means the derivative has rank $r$, so $F$ is a submersion there. The condition is open. It is distinct from global algebraic independence of functions.

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

Let $D={\sum }_p{r}_{p}p\in X^{(d)}$. The derivative of the Abelian sum map is most naturally stated using principal parts. The tangent space of the symmetric product of a curve at $D$ is $H^0({\mathcal{O}}_D(D))$, and the tangent space of the Jacobian variety is $H^0(K_X{)}^{\ast }$. With these identifications,Its dual is restriction of holomorphic one-forms to the length-$d$ divisor scheme. In particular,The last equality is the Riemann-Roch theorem. These formulas apply to repeated points as well as distinct ones.

Here is a direct proof of the repeated-point Abel differential formula. At a multiplicity-$r$ point, choose a local coordinate $t$ with $t(p)=0$. A nearby local divisor is described by the roots $t_1,\dots ,t_r$, or, without making an ordering choice, by their elementary symmetric functions $e_1,\dots ,e_r$:These $e_k$ are smooth local coordinates on the symmetric product of a curve, including at the repeated divisor. Under a first-order deformation, minus the logarithmic variation of this defining polynomial is the principal partThe sign can be checked at $r=1$: moving the point from $0$ to $ϵa$ gives $v=a/t$, and the Abel integral changes by $a\omega (p)$.

Write $\omega =({\sum }_{j\ge 0}{c}_j{t}^j)dt$ and its local primitive as $F(t)={\sum }_{j\ge 0}{c}_j{t}^{j+1}/(j+1)$. The local contribution to the Abelian sum map is ${\sum }_{i}F(t_i)$. Newton's identities, linearized at $e_1=\cdots =e_r=0$, giveSubstituting in the primitive therefore givesSumming over the support proves the theorem. In particular, a multiplicity-$r$ point tests the first $r$ coefficients of a holomorphic one-form, not just its value. The residue pairing between principal parts of order at most $r$ and differential jets through order $r-1$ is perfect. Thus the annihilator of the derivative is precisely the holomorphic one-forms vanishing to order at least $r_p$ at every $p$, namely $H^0(K_X(-D))$.

There is also a sheaf cohomology description that explains its kernel geometrically. The sequencegives a connecting homomorphism $\delta :H^0({\mathcal{O}}_D(D))\to H^1({\mathcal{O}}_X)$. Represent a principal part by meromorphic lifts on coordinate disks; their differences on overlaps give its Čech cohomology class. Under Serre duality, pairing this class with $\omega$ is the sum of the residues of $v_p\omega$, so $\delta$ is the displayed derivative. Exactness shows that its kernel is $H^0({\mathcal{O}}_X(D))/\mathbb{C}$, of dimension $h^0({\mathcal{O}}_X(D))-1$, in agreement with the tangent directions to the complete linear system of a divisor in Solution 2.

Two consequences will be useful below. If $d\ge 2g-1$, then $K_X(-D)$ has negative degree and no nonzero sections, so $u_d$ is a submersion everywhere. Its image is open, and it is also closed because $X^{(d)}$ is compact. Since the Jacobian variety is connected, $u_d$ is surjective. Thus the points $u(p)$ generate the Jacobian variety as a group. Together with Abel's theorem, this identifies the degree-zero Picard group with the Jacobian variety: every line bundle has a meromorphic section and thus a divisor representation, injectivity is Abel's theorem, and surjectivity follows from $u_d(D)-du(p_0)$.

For $g\ge 1$, one can choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$: at each step a surviving nonzero holomorphic one-form has only finitely many zeros, so choose a point where it does not vanish. Their sum $D$ has $h^0(K_X(-D))=1$ and hence $h^0({\mathcal{O}}_X(D))=1$. The derivative has rank $g-1$ there. Consequently $W_{g-1}=u_{g-1}(X^{(g-1)})$ is an irreducible subvariety of dimension $g-1$. This also covers $g=1$, when $X^{(0)}$ and $W_0$ are points.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.

If $L\subset Q$ is a Lagrangian submanifold, its inverse image $\tilde{L}={\pi }^{-1}(L)$ is a smooth circle bundle over $L$. The restricted symplectic form is ${\pi }^{\ast }({\omega }_Q{|}_L)=0$. If $\dim M=2N$, then $\dim Q=2N-2$, $\dim L=N-1$, and $\dim \tilde{L}=N$, proving that $\tilde{L}$ is Lagrangian. This is the Lagrangian lift through circle reduction.

For scalar multiplication on ${\mathbb{C}}^{n+1}$, the generator is $Y={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$. Direct contraction givesChoose $t>0$. Its level is the sphere of radius $R=\sqrt{2t}$ and its quotient is ${\mathbb{CP}}^n$. Complex conjugation on the sphere reverses the ambient symplectic form and descends to an anti-symplectic involution of the quotient. Its fixed set is ${\mathbb{RP}}^n$: a projective point fixed by conjugation has a real representative after a phase change. For two tangent vectors to this fixed set, anti-symplecticity makes the reduced form equal to its negative, so it vanishes. The real dimension is $n$, half the quotient dimension, proving the Lagrangian fixed locus of an anti-symplectic involution property.

The lifted Lagrangian submanifold can be written explicitly asThe only redundancy in this parametrization is the displayed free involution: two nonzero real representatives on the same complex ray differ by a real sign. The quotient embeds smoothly in the sphere, and projection to $[x]\in {\mathbb{RP}}^n$ exhibits its circle bundle structure. Its Lagrangian property follows from the reduction argument, including the case $n=0$.

A smooth right group action is a smooth map $(p,g)\mapsto p\cdot g$ from $P\times G$ to $P$, with $p\cdot e=p$ and $(p\cdot g)\cdot h=p\cdot (gh)$. It is a free action of a group if $p\cdot g=p$ implies $g=e$.

A smooth principal bundle consists of such a free action and a surjective map $\pi :P\to B$ whose fibres are the orbits, together with equivariant local trivializationswhere $(x,g)\cdot h=(x,gh)$. The trivializations are diffeomorphisms; in particular the bundle projection is a submersion.

For two such trivializations, $H={\Phi }_2{\Phi }_1^{-1}$ fixes the base coordinate and is equivariant. Write $H(x,e)=(x,\psi (x))$. Equivariance immediately givesThe function $\psi$ is smooth because $x\mapsto (x,e)$ and $H$ are smooth. This proves the general transition function of a principal bundle formula; the multiplier is on the left, even though the original action is on the right.

For the explicit bundle, represent a point of ${\mathbb{RP}}^3$ by a nonzero vector $w=(w_1,w_2)\in {\mathbb{C}}^2$ modulo nonzero real scaling. The map sends $[w]$ to the complex line $[w_1:w_2]$, so it is well-defined and surjective. Its smoothness follows from the local ratio charts below. Ordinary scalar multiplication by $u\in U(1)$ would not be free, since $u=-1$ acts trivially on real projective classes. Instead defineThe two possible roots differ by a real sign, so give the same class. This is independent of the real representative of $[w]$. Products of chosen roots show the right action law, and the identity acts trivially. Locally on $U(1)$ a smooth square-root branch exists; these local definitions agree projectively and prove global smoothness of the action.

If $[\lambda w]=[w]$, then $\lambda w=rw$ for some nonzero real $r$. Since $w\ne 0$ and $|\lambda |=1$, this forces $\lambda =\pm 1$ and hence $u=1$. The action is free. Two points in the same fibre have representatives $w^{\prime }=cw$ for some $c\in {\mathbb{C}}^{\ast }$; absorb $|c|$ into the real scaling and use $u=(c/|c|{)}^2$. Thus the orbits are exactly the fibres.

Let $N_1=\{[z:1]\}$ and $N_2=\{[1:\zeta ]\}$. Using these base coordinates, defineBoth formulas are invariant under real nonzero scaling, including negative scaling. Their inverses areAgain the sign of the root does not matter. Direct substitution shows they are two-sided inverses. The forward maps are smooth, and local square-root branches prove that the inverses are smooth. Under the right action, each squared phase is multiplied by $u$, so the maps are equivariant. They are therefore genuine principal trivializations of the principal circle bundle on real projective three-space.

On the overlap $z\ne 0$, we have $\zeta =1/z$ andConsequentlyWhen the base point itself, rather than its chart coordinate, is written as $x$, its coordinate remains $x$ in the general transition formula. The opposite transition has reciprocal multiplier. The doubled phase is essential to freeness and is not the ordinary Hopf-bundle phase.

A submersion from a nonempty compact manifold to ${\mathbb{R}}^{\ell }$, $\ell \ge 1$, would be surjective by the preceding result, since the target is connected. Its image would also be compact, contradicting noncompactness of ${\mathbb{R}}^{\ell }$. Thus no such submersion exists. The nonempty-domain qualification excludes the vacuous empty-manifold case.

Coordinate projections are open maps, so the local form of the submersion proves that $f$ is open. If $X$ is compact and nonempty, its continuous image is compact and hence closed in the Hausdorff manifold $Y$. Therefore $f(X)$ is a nonempty set both open and closed in connected $Y$, giving $f(X)=Y$.

Under the given rank hypothesis, the necessary and sufficient local condition isVanishing only at the single point would not suffice: an extension preserves the coordinate Poisson brackets throughout its domain. Necessity follows from the coordinate criterion just proved.

For sufficiency, put $Q_i=a_i$ and $Y_i=X_{Q_i}$. Because the symplectic form is a nondegenerate bilinear form at every point, the independent differentials $d{Q}_i$ give independent fields $Y_1,\dots ,Y_n$. They commute, sinceMoreover $d{Q}_j(Y_i)=\{Q_j,Q_i\}=0$, so they span the tangent spaces of the $n$-dimensional fibres of the submersion $Q$. Choose a local section $s(Q)$ transverse to those fibres. Its joint local flow defines coordinates $(Q,t)$ byThe inverse function theorem makes $\Psi$ a local diffeomorphism, and commuting flows give ${\Psi }_{\ast }{\partial }_{t_i}=Y_i$.

In these coordinates ${\iota }_{{\partial }_{t_i}}{\Psi }^{\ast }{\omega }_0=d{Q}_i$. Each flow preserves the symplectic form, by Cartan's magic formula. Thus its coefficients are independent of $t$, andwhere $\beta =s^{\ast }{\omega }_0$ is a closed differential two-form on the base. On a small ball, the Poincaré lemma gives $\beta =d\alpha$, with $\alpha ={\sum }_i{\alpha }_i(Q)d{Q}_i$. DefineThenConsequently the original coordinates are extended to the required canonical transformation $(Q,P)$, with the prescribed first half $Q=a$. Its Jacobian matrix is invertible by construction. This is the canonical completion of independent commuting functions. For the printed $C^2$ data, the flows and resulting canonical transformation can be taken $C^1$, which is sufficient to preserve the symplectic form; smooth data give smooth coordinates.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

A submersion is a smooth map between manifolds $F:X^n\to Y^m$ for whichis surjective at every $p\in X$.

The local submersion theorem says that around every $p\in X$ there are coordinates $(x^1,\dots ,x^n)$ centered at $p$ and $(y^1,\dots ,y^m)$ centered at $F(p)$ in whichTo prove it, surjectivity lets us choose $m$ source coordinates such that $d{F}^1,\dots ,d{F}^m$ are independent. Complete them by source coordinates $x^{m+1},\dots ,x^n$ and defineThe derivative of $G$ is invertible at $p$, so the inverse function theorem makes $G$ a local coordinate system; in these coordinates $F$ is the displayed projection.

For $q\in Y$, the fiber is locally given byThese are slice coordinates, so $F^{-1}(q)$ is an embedded submanifold of dimension $n-m$.

A vector field $X$ is projectable through a smooth map between manifolds $f:M\to N$ if there is a smooth vector field $Y$ on $N$ with $d{f}_p{X}_p=Y_{f(p)}$ for every $p$. Agreement on fibres is necessary. For a surjective submersion it is also sufficient: smooth local sections of the submersion express $Y$ locally as $df(X)$ and prove its smoothness. An arbitrary non-surjective map may instead give a field only along its image, with extension to $N$ requiring additional choices.

A value is regular when the derivative is surjective at every point of its preimage.

Every value of a submersion is regular, but a map can have some regular values without being a submersion everywhere.

A surjective submersion between Riemannian manifolds is Riemannian when its derivative is a linear isometry from the orthogonal complement of its kernel onto the target tangent space. Its vertical space is the kernel and its horizontal space is the orthogonal complement.

If $F:X^n\to Y^m$ is a submersion at $p$, there are local coordinates in whichEvery fiber of a submersion is consequently an embedded submanifold of codimension $m$.