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A point $x\in X$ is a critical point of a smooth map if $d{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is the image of at least one critical point. A regular value is a point $y\in Y$ for which every point of $f^{-1}(y)$ has surjective differential. A point outside the image is therefore a regular value vacuously. These definitions use rank relative to $\dim Y$, not merely whether a derivative vanishes in chosen coordinates.

For $n\ge 2$, the sphere $S^n$ is simply connected. The lifting criterion for a covering space says that a based map lifts to a covering whenever the image of its fundamental group lies in the subgroup corresponding to that covering. For the universal cover $p:{\mathbb{R}}^2\to T$, the sphere's trivial fundamental group meets this condition. Hence $f:S^n\to T$ has a continuous lift $\tilde{f}:S^n\to {\mathbb{R}}^2$.

Contract the lift linearly to any point $q\in {\mathbb{R}}^2$:This is a continuous homotopy from $f$ to the constant map $p(q)$. Thus every such sphere-to-torus map is null-homotopic. The restriction $n\ge 2$ is essential; circle maps may carry nonzero fundamental-group classes.

The assertion needs $E$ connected. Without that assumption, the disjoint union of two tori mapping by the identity on each component is a covering of the torus and is none of the three listed connected spaces. The correct general statement classifies each connected component; here is the intended connected proof.

The universal cover is ${\mathbb{R}}^2\to {\mathbb{R}}^2/{\mathbb{Z}}^2=T$, and ${\pi }_1(T)={\mathbb{Z}}^2$. The classification of connected coverings of a connected, locally path-connected, semilocally simply connected space identifies $E$ with ${\mathbb{R}}^2/H$ for a subgroup $H\le {\mathbb{Z}}^2$, acting by translations. A subgroup of ${\mathbb{Z}}^2$ is free abelian of rank zero, one or two. For rank zero the quotient is the plane. For rank one, a linear coordinate change sends its generator to a translation in one coordinate, yielding $S^1\times \mathbb{R}$. For rank two, a linear coordinate change sends a lattice basis to the standard basis, yielding a torus. Thus a connected covering is a plane, cylinder or torus. For arbitrary $E$, its covering components are disjoint unions of these types, rather than necessarily one such space.

The symmetric group is transitive on two-element subsets, giving $⟨1,{\chi }_Y⟩=1$. Its diagonal action on $X\times Y$ has two orbits, according to whether the point lies in the pair. Thus $⟨{\chi }_X,{\chi }_Y⟩=2$, and, since ${\chi }_X=1+{\chi }_S$, the standard zero-sum representation $S$ occurs in $\mathbb{C}[Y]$ once. Its irreducibility follows from the double transitivity of the natural symmetric-group action and the corrected statement of part (ii).

For $n\ge 4$, ordered pairs of two-element subsets have three possible intersection sizes, zero, one and two, each forming a single orbit. Part (i) therefore gives $\Vert {\chi }_Y{\Vert }^2=3$. After removing the one copy each of $1$ and $S$, the remaining character has squared norm $3-1-1=1$, so is itself irreducible. HenceFor $n=3$, taking the complement of a pair is an equivariant bijection $Y\to X$. Thus $\mathbb{C}[Y]\cong 1\oplus S$, with no third constituent. Correspondingly the disjoint-pair orbit is absent and the character norm is two.

There is a missing hypothesis in the printed equivalence. A group acting trivially on a two-point set has a one-dimensional zero-sum subspace $S$, hence an irreducible trivial representation, but its action is not doubly transitive. Thus the assertion as written is false. The intended result holds for a transitive action; more generally the trivial two-point action is the only exceptional case under $|X|>1$.

For the intended proof, decompose $\mathbb{C}[X]=1\oplus S$ by separating constant functions from functions with zero sum. If the action is transitive, $⟨1,{\chi }_X⟩=1$. HenceOver the complex numbers, complete reducibility and character orthogonality identify this norm as the sum of squared irreducible multiplicities; it is one precisely when $S$ is irreducible. By part (i), $\Vert {\chi }_X{\Vert }^2$ counts orbits on $X\times X$. Transitivity gives one diagonal orbit, and the norm equals two exactly when all off-diagonal ordered pairs form one orbit, which is double transitivity. This proves both implications with the missing transitivity condition restored.

To check the exact exception, if the action has $r>1$ orbits, then $S$ contains an invariant subspace of dimension $r-1$, from orbitwise constant functions with total sum zero. If $S$ is irreducible this makes $S$ the one-dimensional trivial representation, so $|X|-1=1$ and $|X|=2$. Nontransitivity on two points means both are fixed. Therefore irreducibility is equivalent to double transitivity, except for the trivial two-point action.

The character of a permutation representation at $g$ equals the number of points fixed by $g$, because a permuted basis vector contributes one to the trace exactly when it is fixed. Thus the character inner product isTo identify the last average, count pairs $(g,z)$ with $gz=z$. Each orbit contributes $|G|$, since its number of elements times their common stabilizer size is $|G|$ by the orbit-stabilizer theorem. ThereforeThe conjugation in the usual character inner product has no effect here because the permutation characters are real-valued.

Measure $\theta$ from the downward vertical, so the bob is at $(x+l\sin \theta ,-l\cos \theta )$. Its squared speed is ${\dot{x}}^2+2l\dot{x}\dot{\theta }\cos \theta +l^2{\dot{\theta }}^2$. Up to a constant potential term, the Lagrangian isThe Euler-Lagrange equations giveAt the downward stable equilibrium, linearization gives $(M+m)\ddot{x}+ml\ddot{\theta }=0$ and $l\ddot{\theta }+\ddot{x}+g\theta =0$. Eliminating $\ddot{x}$ yieldsThere is also a zero-frequency translational mode because the rail fixes no preferred pivot position.

If an external force imposes $\ddot{x}=A$ constantly, the bob's angular equation becomes $l\ddot{\theta }+A\cos \theta +g\sin \theta =0$. Its stable equilibrium therefore hason the branch near the downward vertical. The displaced bob leans opposite the acceleration. The solution differing by $\pi$ is unstable, corresponding to alignment opposite the effective gravitational field in the accelerating frame.

Consider a smooth variation $q^a(t)+ϵ{\eta }^a(t)$ with ${\eta }^a(t_1)={\eta }^a(t_2)=0$. Differentiating the action at zero and integrating the velocity term by parts givesThe boundary term vanishes because the endpoints are fixed. Independent compactly supported variations and the fundamental lemma of the calculus of variations make every bracket zero. Thus the Euler-Lagrange equations areThe required principle is stationarity of the action; an extremal need not be a strict minimum.

If the weak-rate coefficient is reduced, equality with the expansion rate occurs at a higher temperature, since ${\Gamma }_W/H$ scales as that coefficient times $T^3$. The freeze-out temperature and interaction strength relation is $T_f\propto C_W^{-1/3}$. The frozen neutron-to-proton ratio is then closer to one, so the primordial helium mass fraction increases, potentially approaching unity if nearly equal numbers survive to nucleosynthesis. If the same weaker interaction also slows free-neutron decay, it reinforces this trend; changing conversion rates alone would instead require a separate treatment of the decay interval.

Less primordial hydrogen means less hydrogen fuel and, for otherwise comparable stellar physics, helium enrichment raises the mean molecular weight, tending to make a fixed-mass star hotter and more luminous and to shorten its hydrogen-burning lifetime. A genuinely weaker weak interaction also changes proton-proton fusion itself, so a quantitative lifetime cannot be inferred from composition alone. Much less hydrogen and fewer long-lived suitable stars would alter planet formation, water availability and the timescales needed for life on Earth. These are physical implications of the changed initial composition, not a numerical prediction of habitability from the helium fraction alone.

The weak-conversion rate per expansion time falls as $T^3$. Below the stated threshold it becomes less than one, so neutron-proton interconversion cannot track the rapidly changing equilibrium ratio: cosmological weak freeze-out occurs. Free neutrons subsequently decay, reducing their abundance until nuclei can survive photodissociation. Once the deuterium bottleneck is passed, nuclear reactions rapidly assemble neutrons into the strongly bound helium-4 nucleus. The short available time, low density and missing stable mass-five and mass-eight nuclei impede extensive production of heavier nuclei, leaving nearly all surviving neutrons in helium.

If $r=n_n/n_p$ just before helium formation, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$. Neglecting binding-energy corrections to nuclear masses,The remaining protons mainly become hydrogen. Counting all four helium nucleons, rather than only the captured neutrons, is essential to this mass fraction.

For dilute nonrelativistic nucleons in thermal equilibrium, their Maxwell-Boltzmann distribution gives $n_j=g_j(m_{j}kT/2\pi {\hslash }^2{)}^{3/2}\exp [({\mu }_j-m_j{c}^2)/kT]$. The neutron and proton spin degeneracies are equal, and their mass prefactor ratio is nearly one. ThusWeak reaction equilibrium imposes ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Charge neutrality in an electron-positron plasma, together with the small baryon asymmetry, makes ${\mu }_e/kT$ very small. Assuming negligible neutrino lepton asymmetry also makes ${\mu }_{{\nu }_e}/kT$ small. Hence the difference, rather than the absolute baryon chemical potentials, is negligible, yieldingA large neutrino chemical potential would invalidate this last approximation; it is an assumption about the lepton asymmetry, not a general rule that baryonic chemical potentials vanish.

Put $\eta =({\gamma }_2-a)/({\gamma }_1-a)$. The strict inequality gives $|\eta -1|<1$, so $\mathrm{Re}\eta >0$. This is a closed piecewise-smooth curve in the right half-plane, and part (b) gives $n(\eta ,0)=0$. Since ${\gamma }_2-a=({\gamma }_1-a)\eta$, part (a) yieldsThis is the dominated perturbation preserves winding number principle, here derived by logarithmic differentiation.

On the right half-plane a single-valued holomorphic logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$, with $-\pi /2<\mathrm{Arg}z<\pi /2$, and its derivative is $1/z$. ThusThis proves the conclusion directly from the integral definition of the winding number, without appealing to its homotopy invariance.

Where both derivatives exist, the ordinary product rule givesNeither denominator vanishes. Integrate over a common piecewise-smooth subdivision and use the contour characterization of the winding number:

For $n\ge 2$, use the unbiased residual variance $s^2=\mathrm{RSS}/(n-1)=n{\hat{\sigma }}^2/(n-1)$. Under the null hypothesis, $(\hat{\beta }-{\beta }_0)\sqrt{S_{xx}}/\sigma$ is standard normal and independent of $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-1}^2$. HenceFor a chosen significance level $\alpha$, reject the null in favor of the two-sided alternative exactly when $|T|>t_{n-1,1-\alpha /2}$, the stated upper quantile of the Student t-distribution. The exact two-sided p-value is $2P(t_{n-1}\ge |T_{\mathrm{obs}}|)$. This hypothesis test does not substitute the biased MLE for $s^2$ without correcting its factor and does not use a normal critical value when variance is unknown.

Let $x$ and $ϵ$ denote the design and error vectors, and put $P=x{x}^T/S_{xx}$. The residual vector is $r=(I-P)ϵ$, while $\hat{\beta }-\beta =x^Tϵ/S_{xx}$. Since $(I-P)x=0$,These are jointly Gaussian linear functions of $ϵ$, so the entire residual vector is independent of $\hat{\beta }$, not merely uncorrelated with it. Choose an orthonormal basis whose first vector is $x/\sqrt{S_{xx}}$. In that basis $ϵ/\sigma$ has independent standard normal coordinates, and the residual keeps exactly the remaining $n-1$ coordinates. ThereforeTogether with part (ii), this specifies the joint distribution as the product of the slope's normal law and the scaled chi-squared distribution. In particular $E{\hat{\sigma }}^2=(n-1){\sigma }^2/n$: the variance maximum-likelihood estimator is biased downward.

The least-squares estimator satisfiesA linear combination of independent normal random variables is normal; its mean is $\beta$ and its variance is ${\sigma }^2{\sum }_i{x}_i^2/S_{xx}^2$. Hence

Assume $S_{xx}={\sum }_i{x}_i^2>0$, which is necessary to identify the slope in linear regression through the origin. The joint normal likelihood has logarithm, apart from a constant,Complete the square in $\beta$. With $\hat{\beta }={\sum }_i{x}_i{Y}_i/S_{xx}$ and $\mathrm{RSS}={\sum }_i(Y_i-\hat{\beta }{x}_i{)}^2$,The maximum likelihood estimation therefore first minimizes at $\hat{\beta }$; differentiating in ${\sigma }^2$ then maximizes atThis is a genuine finite maximum when $\mathrm{RSS}>0$. If all $x_i=0$, the slope is unidentifiable. If $\mathrm{RSS}=0$, the likelihood is unbounded as ${\sigma }^2\downarrow 0$, so no positive-variance maximum exists. For nonzero design and $n\ge 2$, zero residual has probability zero under the stated positive-variance model.

The symmetric coefficient matrix of the quadratic form isThus $\boxed{q_a\text{ is nondegenerate if and only if }a\ne 1.}$ To obtain its signature, put $u=(y+z)/2$ and $v=(y-z)/2$. Completing the square yieldsThe transformation from $(x,y,z)$ to $(u+x,v,x)$ is invertible. The inertia of a bilinear form is consequentlyIf signature means the signed difference $n_{+}-n_{-}$, its value is $+1$ for $a>1$ and $-1$ for $a<1$. At $a=1$ the diagonal form has one zero direction, confirming degeneracy.

Apply the Gram-Schmidt process explicitly. Start with $f_1=e_1/\Vert e_1\Vert$, and for $i\ge 2$ setAssume inductively that the previous $f_j$ form an orthonormal basis for the span of $e_1,\dots ,e_{i-1}$. Inner products give $⟨v_i,f_j⟩=0$ for every $j<i$. Moreover $v_i\ne 0$: otherwise $e_i$ would lie in the span of its predecessors, contradicting that the $e_j$ form a basis. Thus $f_i$ is well defined and has unit norm. The definition expresses $f_i$ in the span of $e_1,\dots ,e_i$, and also expresses $e_i$ in the span of $f_1,\dots ,f_i$. The two spans therefore agree at each stage. Induction proves the required orthonormal, span-preserving basis.