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False. Take $A=\{0\}$ and $B=C=\{0,1\}$, with $f(0)=0$ and $g={\mathrm{id}}_B$. Then $f$ is an injection and $g$ is a surjection, but the function composition $g\circ f$ has image of a function $\{0\}$, so it is not a surjection onto $C$.

Use $D=(d/dt{)}_S$, $D^{\prime }=(d/dt{)}_{S^{\prime }}$, and ${\mathbf{v}}^{\prime }=D^{\prime }\mathbf{r}$. By the rotating-frame derivative formula, $D\mathbf{r}={\mathbf{v}}^{\prime }+\mathbf{\omega }\times \mathbf{r}$. Apply the product rule for the cross product and the same rotating-frame derivative formula again:We have $D{\mathbf{v}}^{\prime }=D^{\prime }{\mathbf{v}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$ and $D\mathbf{\omega }=D^{\prime }\mathbf{\omega }$, since $\mathbf{\omega }\times \mathbf{\omega }=0$. ThereforeRearranging this identity into an equation of motion in a rotating frame produces the apparent Coriolis acceleration, Euler acceleration and centrifugal acceleration, with signs opposite to their corresponding terms on the right of this inertial acceleration identity. No translational term is needed because the two origins coincide.

Let ${\mathbf{e}}_1^{\prime }(t),{\mathbf{e}}_2^{\prime }(t),{\mathbf{e}}_3^{\prime }(t)$ be an orthonormal basis fixed in the rotating reference frame $S^{\prime }$. A small rotation through $\mathbf{\omega }dt$ changes each basis vector by $\mathbf{\omega }\times {\mathbf{e}}_j^{\prime }dt$, so its derivative in the inertial frame isWrite the arbitrary vector as $\mathbf{a}={\sum }_j{a}_j(t){\mathbf{e}}_j^{\prime }(t)$. Differentiating the components and the moving basis separately givesThis proves the rotating-frame derivative formula, including when the angular velocity varies with time.

Choose the gravitational potential energy to vanish at infinity. The total mechanical energy isPut $\gamma =GM/h^2$ and $\phi =\theta -{\theta }_0$. From part (i), $u=\gamma (1+e\cos \phi )$ and $u^{\prime }=-\gamma e\sin \phi$. Consequentlyand the angular terms cancel against the gravitational potential energy:For a nonradial Kepler orbit, negative, zero and positive mechanical energy correspond respectively to $e<1$, $e=1$ and $e>1$.

Multiplying the tangential equation of motion by $r$ givesTherefore $h=r^2\dot{\theta }$ is constant. It is the signed specific angular momentum; reverse the angular orientation if necessary to take $h>0$ for a nonradial orbit. Set $u=1/r$ and denote differentiation with respect to $\theta$ by a prime. ThenSubstitution into the radial equation of motion and division by $-h^2{u}^2$ give the Binet equationIts general solution is $u=GM/h^2+A\cos \theta +B\sin \theta$. Write $A=(GM/h^2)e\cos {\theta }_0$ and $B=(GM/h^2)e\sin {\theta }_0$, where $e\ge 0$. This produces the Kepler orbitHere $e$ is the orbital eccentricity, and ${\theta }_0$ points toward periapsis when $e>0$. Only portions with $u>0$ describe positive radial distances. The expression assumes $h\ne 0$; a zero-angular momentum trajectory is radial and must be treated separately.

The derivative matrixis symmetric in $j,k$. Contracting it with the antisymmetric Levi-Civita symbol gives the curlDirect differentiation of the scalar potential givesso $\mathbf{F}=\nabla \varphi$. To identify the level sets, use the cross product identityIf $\mathbf{\omega }\ne 0$, write $\mathbf{x}={\mathbf{x}}_{\parallel }+{\mathbf{x}}_{\perp }$ relative to its direction. Then $\varphi =-|\mathbf{\omega }{|}^2|{\mathbf{x}}_{\perp }{|}^2/2$. Thus each negative level $\varphi =c$ is a circular cylinder with axis $\mathbb{R}\mathbf{\omega }$ and radius $\sqrt{-2c}/|\mathbf{\omega }|$. The zero level is the axis itself; positive levels are empty. If $\mathbf{\omega }=0$, the scalar potential is identically zero, so its zero level is all of space and every other level is empty.

The vector triple product removes the nested cross products:Since $\mathbf{\omega }$ is constant, the divergence isThe cube has volume $2^3=8$. With the outward unit normal, the divergence theorem therefore gives

The assertion is true over the complex numbers. It suffices to generate every nonzero scaling using translations and $I(z)=1/z$, since these together with scalings already generate the Möbius group.

For $s\ne 0$, compose the maps rightmost first. Away from the intermediate poles, the successive values areHenceFor every $\lambda \ne 0$, choose a complex square root $s$ with $s^2=-\lambda$. This gives $D_{\lambda }$ and proves Möbius scaling generated by translations and reciprocal inversion. Equality of the rational maps extends the identity to all points of the Riemann sphere, including the intermediate exceptional values.

The assertion is false. Let $D_{\lambda }(z)=\lambda z$ and $I(z)=1/z$. Their relations areMoving each inversion to the right reduces every word to $D_{\lambda }{I}^{ϵ}$ with $ϵ=0$ or $1$. Thus the subgroup generated by complex scalings and reciprocal inversion consists only of $z\mapsto \lambda z$ and $z\mapsto \lambda /z$. Of these, only the first form fixes zero.

The Möbius transformationfixes zero but is not a scaling: it sends infinity to $1$ instead of infinity. It therefore cannot belong to the subgroup generated by scalings and reciprocal inversion.

The assertion is true. A Möbius map fixes infinity if and only if $c=0$: if $c\ne 0$, its value there is the finite number $a/c$. For $c=0$, the decomposition above uses only a nonzero complex scaling and a translation. These maps form the Affine subgroup of the Möbius group.

Assume $H=\ker \theta$ for a group homomorphism $\theta :G\to K$. For $h\in H$ and $g\in G$,Thus $gH{g}^{-1}\subseteq H$. Apply the same inclusion with $g^{-1}$ to obtain the reverse inclusion, giving $gH{g}^{-1}=H$. Hence every kernel is a normal subgroup, proving the reverse implication and the required equivalence.

Assume $H$ is a normal subgroup. Take the quotient group $K=G/H$ and the natural projection $\theta (g)=gH$. The well-defined multiplication constructed above gives $\theta (gk)=\theta (g)\theta (k)$, so this is a group homomorphism. Its identity is the coset $H$, and $\theta (g)=H$ exactly when $g\in H$. Thereforewhich proves the implication from normality to being the kernel of a group homomorphism.

Divide the velocity by the speed calculated above. The unit tangent vector isIts derivative isBecause $ds/dt=\sqrt{2}(1+t^2)>0$, the principal normal vector has the direction of $T^{\prime }(t)$, givingThis is a unit vector, and $N\cdot e_z=0$ for every real $t$. The curvature of a space curve is $|T^{\prime }|/(ds/dt)=1/(1+t^2{)}^2>0$, so the principal normal vector is defined everywhere on this curve. The same perpendicularity follows from constant tangent component forces perpendicular principal normals: $T\cdot e_z=1/\sqrt{2}$ is constant, and differentiating this identity forces the normal component along $e_z$ to vanish.

Differentiate the parametrized curve to obtain ${\mathbf{x}}^{\prime }(t)=(1-t^2,2t,1+t^2)$. Its speed obeysSince $1+t^2>0$, the arc length is

The equality with part (i) follows from Green's theorem. For the coefficients $P=-1$ and $Q=x^2/(2y)$,They are smooth on a neighborhood of this region, since $y>0$. Thus for the positively oriented boundary,The term $-dx$ itself has zero line integral around the closed curve, although it contributes to the four individual edge integrals.

Take the usual positive, counterclockwise boundary orientation; the printed question does not specify one, and reversing it negates the answer. Write the successive corner points as $P=(1,1)$, $Q=(\sqrt{2},1/\sqrt{2})$, $R=(2,1)$, $S=(\sqrt{2},\sqrt{2})$. Traverse $P\to Q\to R\to S\to P$, keeping the region on the left.

On $PQ$, use $y=1/x$ with $1\le x\le \sqrt{2}$. Here $dy=-dx/x^2$, so the line integral isOn $QR$, use $y=x/2$ with $\sqrt{2}\le x\le 2$:On $RS$, use $y=2/x$ with $x$ decreasing from $2$ to $\sqrt{2}$:Finally, on $SP$, use $y=x$ with $x$ decreasing from $\sqrt{2}$ to $1$:Adding all four line integrals gives ${\oint }_C(x^2/(2y))dy-dx=1/2$ for the positive orientation, or $-1/2$ for the reverse orientation.

The inequalities $y\le x\le 2y$ force $y\ge 0$, and the reciprocal bounds exclude zero. Thus the whole region lies in the positive quadrant. Introduce ratio-product coordinatesThe region becomes the rectangle $1\le u\le 2$, $1\le v\le 2$. The inverse map and its Jacobian determinant areThe change of variables formula then gives