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

The Riemann-Hurwitz formula for a nonconstant degree-$n$ morphism
$f:C\to C^{\prime }$ of smooth projective connected curves in characteristic zero iswhere $R_f$ is the effective ramification divisor. If $g(C^{\prime })\ge 2$ and
$g(C)<g(C^{\prime })$, the right side is at least $2g(C^{\prime })-2>2g(C)-2$, impossible. If $g(C^{\prime })=1$, then $g(C)=0$ and the left side is $-2$ while the right side is nonnegative. Thus every morphism is constant.

Choose $p\in C_d$ and project from $p$ to the pencil of lines through it. Bézout's theorem says that a general such line meets $C_d$ in $p$ plus $d-1$ further points, so the resolved projectionhas degree $d-1$. Since the genus of a smooth plane curve is
$g=(d-1)(d-2)/2$, Riemann--Hurwitz givesEvery branch point receives at least one ramification point, hence

The adjunction formula givesSince $\mathrm{deg}D=d$ and $\mathrm{deg}{K}_{C_d}=d(d-3)$, linear equivalence
$D\sim K_{C_d}$ would force $d=d(d-3)$, hence $d=4$. It is therefore impossible for $d\ge 5$.

For a smooth plane quartic, projection from a point of the curve gives a degree-three map, so its gonality is at most three. Its genus is three and
$K_C\sim D$, so its plane embedding is the canonical map. A genus-three curve with a degree-two map to ${\mathbb{P}}^1$ would be hyperelliptic, and its canonical map would factor through that double cover and map onto a conic rather than embed the curve. Thus no degree-two map exists, and a degree-one map is excluded by the genus. The quartic's gonality isFor a genus-one curve, Riemann--Roch supplies a degree-two map to ${\mathbb{P}}^1$, while degree one would make it isomorphic to ${\mathbb{P}}^1$. Its gonality is therefore

A functional $p:X\to \mathbb{R}$ is sublinear when

Any extension to $\tilde{M}=M+\mathbb{R}x$ must have the formChoose $c$ satisfyingThis interval is nonempty: for $y,z\in M$,which rearranges to the required lower bound being at most the upper bound. For $t>0$, positive homogeneity reduces
$\tilde{\ell }(y+tx)\le p(y+tx)$ to the upper inequality after replacing $y$ by $y/t$; for $t<0$ it reduces to the lower inequality. The case $t=0$ is the hypothesis on $\ell$. Thus this choice gives the required dominated linear extension.

The dominated form of the Hahn-Banach theorem states that a linear functional on a subspace, bounded above by a sublinear functional, extends linearly to the whole real vector space while retaining that bound.

Let $M=\mathrm{span}\{z_1,\dots ,z_n\}$. Its coordinate mapsare continuous because every linear map on a finite-dimensional normed space is continuous. Hahn--Banach extends each to
${\ell }_j\in Z^{\prime }$ without increasing its norm, and then

For an arbitrary finite-dimensional subspace $M$, choose a basis $z_1,\dots ,z_n$ and these extended coordinate functionals. Thenis closed. Every $z\in Z$ has the decompositionApplying every ${\ell }_j$ shows $M\cap N=\{0\}$, so

The Snake lemma associates to a commutative diagram with exact rows an exact sequenceFor $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexesThe resulting connecting maps splice the kernels modulo boundaries into the Mayer--Vietoris sequence

Now let $K$ satisfy the stated simplicial pseudomanifold conditions. In an $n$-cycle, cancellation at a common $(n-1)$-face determines the coefficient of either incident $n$-simplex from the other, up to the orientation sign. Connectivity through such faces therefore determines every top-simplex coefficient from one integer. If the signs are globally compatible, their oriented sum is a cycle and generates $H_n(K)\cong \mathbb{Z}$; if they are inconsistent, that integer must be zero and $H_n(K)=0$.

Let $x$ be the resulting fundamental class, represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Thenand the Mayer--Vietoris boundary isIt is nonzero exactly when both $L_1$ and $L_2$ contain top-dimensional simplices; if one side contains none, the fundamental cycle already lies in the other side, while if both do, connectedness forces a nonempty interface and its oriented boundary represents a nonzero class.

Finally take $K\cong S^3$, $L_1\cong S^1\times D^2$, and
$L_1\cap L_2\cong T^2$. The preceding boundaryis a nonzero map between copies of $\mathbb{Z}$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then givesThe next part of the sequence isSince $H_1(L_1)\cong \mathbb{Z}$, it follows that
$H_1(L_2)\cong \mathbb{Z}$. The degree-zero sequence makes $L_2$ connected. Therefore

Write an integral binary quadratic form as $[a,b,c]=a{x}^2+bxy+c{y}^2$. It is positive definite when $a>0$ and its discriminant $D=b^2-4ac$ is negative. Two forms are properly equivalent when one is obtained from the other by a change of variables in ${\mathrm{SL}}_2(\mathbb{Z})$. A reduced positive definite binary quadratic form satisfieswith $b\ge 0$ when $|b|=a$ or $a=c$.

To prove reduction, choose in a proper equivalence class a form whose positive leading coefficient $a$ is minimal. Replacing $x$ by $x+ny$ changes $b$ by $2an$, so choose $n$ to arrange $|b|\le a$. If the resulting $c<a$, the determinant-one substitution $(x,y)\mapsto (-y,x)$ would put $c$ in the leading position, contradicting minimality. Thus $a\le c$. Sign changes on the boundary give the stated convention. Hence every positive definite form is properly equivalent to a reduced one.

A unimodular integral substitution is a bijection of ${\mathbb{Z}}^2$, so properly equivalent forms represent exactly the same integers.

Both given forms have discriminant $-23$. Reduction givesThe two reduced forms represent the same integers becausean integral change of variables of determinant $-1$. They are not properly equivalent: the uniqueness theorem for reduced positive definite forms says that each proper class has one reduced representative, and $[2,1,3]\ne [2,-1,3]$. Therefore the original forms represent the same integers but are not equivalent in the required proper sense.

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.