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Fix $t_0\in [a,b]$ and defineBecause $\gamma$ is a regular curve, $s^{\prime }(t)=\Vert {\gamma }^{\prime }(t)\Vert >0$. The inverse function theorem gives a smooth inverse $t=t(s)$ onto the corresponding interval. For $\tilde{\gamma }(s)=\gamma (t(s))$,Thus the arc-length parametrization exists.

At a point with $xy\ne 0$, the determinant of the derivative isThe inverse function theorem therefore supplies neighbourhoods on which $f$ has a continuously differentiable, hence differentiable, inverse. Concretely, if $(u,v)=f(x,y)$, the local inverse is the restriction ofwhich is continuously differentiable near the image whenever $y\ne 0$.

Because the bilinear form from part (ii) is invariant, $\rho (SU(2))\subseteq O(3)$. The mapis continuous. Since $SU(2)$ is connected and $\det \rho (I_2)=1$, it is identically one. Hence

To find the kernel, suppose $U$ commutes with every traceless skew-Hermitian matrix. Commutation withforces $U$ to be diagonal. Commutation withthen forces its two diagonal entries to agree. Thus $U=z{I}_2$. Since $U\in SU(2)$,and therefore

It remains to prove surjectivity. The derived representation at the identity isIf this vanishes for every $A$ in the SU(2) Lie algebra, the same commutant calculation makes $X$ scalar; tracelessness then gives $X=0$. Thus $d{\rho }_{I_2}$ is injective. Its domain and codomain are both three-dimensional real Lie algebras, so it is an isomorphismThe inverse function theorem now shows that the image of $\rho$ contains a neighbourhood of the identity in $SO(3)$. Therefore the image is an open subgroup. The special orthogonal group $SO(3)$ is connected, since every rotation can be deformed continuously to the identity by reducing its rotation angle. A connected topological group has no proper open subgroup, soTogether with the kernel calculation, this is the Adjoint double cover from SU(2) to SO(3).

A function $f:{\mathbb{R}}^m\to {\mathbb{R}}^r$ is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^r$ such thatThe map $L$, which is unique, is the derivative $f^{\prime }(x)$.

Expanding in a matrix increment $H$ givesTherefore $p$ is differentiable everywhere and

The inverse function theorem states that if $f$ is continuously differentiable near $a$ and $f^{\prime }(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivativeIn the normalized case $f(0)=0$ and $f^{\prime }(0)=I$, continuity of $f^{\prime }$ gives a closed ball ${\overline{B}}_r(0)$ on whichFor small $y$, define $T_y(x)=x-f(x)+y$. The mean value inequality makes $T_y$ a contraction mapping with constant $q$. If $\Vert y\Vert \le (1-q)r$, thenso $T_y$ maps the ball into itself. The contraction mapping theorem gives a unique fixed point $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,so $\Vert x_y-x_z\Vert \le \Vert y-z\Vert /(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.

For the polynomial map in the question,The derivative is invertible, so the inverse function theorem supplies some $ϵ>0$ and a continuously differentiable $q:D_{ϵ}(I)\to {\mathcal{M}}_n$ such thatThere can be no inverse on all of ${\mathcal{M}}_n$, becauseso $p$ is not injective.

This can be false. The functionis a differentiable bijection of $\mathbb{R}$, buthas an unbounded difference quotient at zero and is not differentiable there. The inverse function theorem requires the additional local condition $f^{\prime }(x)\ne 0$.

Outside the finite set of branch points, the derivative is nonzero. The inverse function theorem supplies evenly covered neighbourhoods, and compactness of the Riemann sphere makes every fibre finite, so the restriction is a covering map.

Lift a loop $\gamma$ from each point of the fibre over $z_0$; its endpoints define a permutation ${\sigma }_{\gamma }$. The homotopy lifting property shows that this permutation depends only on the based homotopy class. The punctured source is connected, so a path between any two points in the fibre projects to a loop whose lift joins them. Hence the monodromy group acts transitively.

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

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
$\boxed{X_0\text{ is a one-dimensional manifold}}$.

The Arzela-Ascoli theorem makes the family of iterates relatively compact in $C([0,T_1])$. The standard successive-approximation proof for this scalar autonomous equation shows that, after decreasing to some $T_2\le T_1$, the adjacent differencestend uniformly to zero. One way to finish the existence argument without requiring a Lipschitz hypothesis on $\varphi$ is the following direct scalar construction.

If $\varphi (0)=0$, the function $f(t)=0$ is already a solution. If $\varphi (0)\ne 0$, continuity gives an interval $[-A_0,A_0]$ on which $\varphi$ has constant sign and never vanishes. DefineThen $F$ is continuously differentiable with $F^{\prime }(0)\ne 0$. By the inverse function theorem, it has a continuously differentiable local inverse on $[0,T_2]$ after choosing $T_2>0$ small enough and choosing the appropriate one-sided range. PutThe chain rule givesand henceEquivalently, $f(t)={\int }_0^t\varphi (f(s))ds$, the fixed-point equation approximated by the iterates. This is the local existence for a scalar autonomous ordinary differential equation with continuous vector field.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Frechet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

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 $n-m$.

For $v$ near $0\in T_{p}M$, let ${\gamma }_v$ be the unique geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The exponential map is ${\exp }_p(v)={\gamma }_v(1)$.

The constant initial velocity $0$ gives ${\exp }_p(0)=p$. Varying the initial velocity through $sv$ yields ${\exp }_p(sv)={\gamma }_v(s)$, soThus $(d{\exp }_p{)}_0$ is the identity. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighborhood of $0$ onto a neighborhood of $p$. Coordinates from an orthonormal basis of $T_{p}M$ are the geodesic normal coordinates.

A map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$ if there is a linear map $L$ such thatthen $Df{|}_x=L$. The inverse function theorem says that if $f$ is continuously differentiable near $x$ and $Df{|}_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.

Since $F(A)=A^{T}A$ is polynomial,Thus $\ker DF{|}_I=T$, the space of skew-symmetric matrices.

DefineThen $D\Phi {|}_I(H)=2H$, so the inverse function theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi :U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^{T}A-I$ and its skew part is $A-A^T$. Consequentlyand therefore $\Phi (U\cap \mathcal{O})=V\cap T$.

For any $R\in \mathcal{O}$, left multiplication $A\mapsto R^{T}A$ is a homeomorphism preserving $\mathcal{O}$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal{O}$ homeomorphic to an open subset of $T$.

A horizontal lift of $\gamma$ from $p$ is a curve $\tilde{\gamma }$ in $E$ projecting to $\gamma$, starting at $p$, and tangent to the horizontal distribution of the connection. In a local frame write $\tilde{\gamma }(t)=(\gamma (t),v(t))$. Horizontality is the linear ordinary differential equationwhose initial-value theorem gives local existence and uniqueness; successive trivializations continue the lift.

A geodesic satisfies ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$, andfor the geodesic with initial velocity $v$. Since $d({\exp }_p{)}_0=1$, the inverse function theorem makes ${\exp }_p$ a diffeomorphism near zero; its inverse gives normal coordinates. A geodesic sphere is ${\exp }_p(\{v:|v|=r\})$ inside such a normal neighborhood.

The Gauss lemma statesFor the variation ${\gamma }_s(t)={\exp }_p(t(v+sw))$, let $J={\partial }_s{\gamma }_s{|}_{s=0}$. The coordinate vector fields commute, so metric compatibility and constant geodesic speed giveSince $J(0)=0$, evaluation at $t=1$ proves the formula. In particular radial and spherical directions are orthogonal.

A gauge condition $F(x)=0$ is regular at a representative $x_0$ if its derivative along the gauge orbit is an invertible matrix. The inverse function theorem then makes the gauge-condition values valid local coordinates along that orbit. A unique zero without this regularity is insufficient: the translation action on the line with $F(x)=x^3$ has a unique zero but a zero derivative there. Multiple regular intersections give a Gribov ambiguity rather than a globally unique slice.

A Lagrangian is regular when its velocity Hessian matrix is invertible. The inverse function theorem then makes the fiber derivative $p_i=L_{v^i}$ locally invertible, allowing a local Legendre transform in mechanics to a Hamiltonian. If that fiber derivative is globally invertible, the Lagrangian is called a hyperregular Lagrangian. Regularity alone does not assert global invertibility, and singular Lagrangians may give constrained rather than ordinary unconstrained Hamilton's equations.

Near a point where a smooth vector field is nonzero, there are coordinates in which it is $\partial /\partial x_1$. Flow a transverse coordinate slice by the local flow; the inverse function theorem makes the resulting map a coordinate chart. This is the first step of the coordinate proof of the Frobenius theorem.