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Suppose first that $F:D^n\to X$ extends $f$. Regarding $S^{n-1}$ as the boundary of the unit disc,is a homotopy from $f$ to the constant map with value $F(0)$. Thus $f$ is null-homotopic. Conversely, if $H:S^{n-1}\times [0,1]\to X$ is a homotopy from $f$ to a constant, then $H$ is constant on $S^{n-1}\times \{1\}$ and hence descends to the quotientwhich is the cone on $S^{n-1}$ and is homeomorphic to $D^n$. The descended map extends $f$. This proves the extension-null-homotopy criterion for a sphere.

A universal cover of $X$ is a covering map $p:\tilde{X}\to X$ whose total space is path-connected and simply connected. Let $p_1:({\tilde{X}}_1,{\tilde{x}}_1)\to (X,x)$ and $p_2:({\tilde{X}}_2,{\tilde{x}}_2)\to (X,x)$ be two universal covers. The lifting criterion for a covering space applies becauseso $p_1$ has a based lift $h:{\tilde{X}}_1\to {\tilde{X}}_2$ satisfying $p_{2}h=p_1$. Similarly there is a based lift $k:{\tilde{X}}_2\to {\tilde{X}}_1$ of $p_2$. Both $kh$ and the identity are lifts of $p_1$ that agree at the base point, so uniqueness of lifts gives $kh=1$. Likewise $hk=1$. Hence $h$ is a homeomorphism. This is the uniqueness of a universal covering space.

Now let $p:\tilde{X}\to X$ be universal with $\tilde{X}$ contractible. If an extension $F:|K|\to X$ exists, functoriality of the fundamental group givesThus the required factorization holds with $\Phi =F_{\ast }$.

Conversely, suppose $f_{\ast }=\Phi i_{\ast }$. We extend over the simplices of $K$ by dimension. The boundary of every 2-simplex is a loop in $|K^1|$ that becomes null-homotopic in $|K|$, so its class lies in $\ker i_{\ast }$. The factorization givesHence $f{|}_{\partial \sigma }$ is null-homotopic in $X$ and extends over the disc $|\sigma |$ by the first part. Since distinct 2-simplices meet along the already fixed 1-skeleton, these extensions combine to a map on $|K^2|$.

Inductively suppose the map is defined on $|K^{m-1}|$ for $m\ge 3$. For an $m$-simplex $\sigma$, its boundary is $S^{m-1}$, which is simply connected. The boundary map therefore lifts through $p$ by the covering-space lifting criterion. Its lift into the contractible space $\tilde{X}$ is null-homotopic, so the boundary map itself is null-homotopic and extends over $|\sigma |$. Extending over every $m$-simplex and then every skeleton constructs $F:|K|\to X$. The weak topology on a simplicial complex makes the cellwise map continuous. This proves the extension criterion into a space with contractible universal cover.

The Lagrange theorem for polynomial congruences says that if $p$ is prime and $f\in \mathbb{Z}[X]$ has degree $d$ with at least one coefficient not divisible by $p$, thenhas at most $d$ incongruent solutions modulo $p$.

The Chinese remainder theorem states that for pairwise coprime $m_1,\dots ,m_r$, every systemhas exactly one solution modulo $M={\prod }_i{m}_i$. For two moduli, choose $u,v$ with $um+vn=1$ by Bezout identity. Thenis congruent to $a$ modulo $m$ and to $b$ modulo $n$. If $x,y$ are two solutions, both $m$ and $n$ divide $x-y$; coprimality makes $mn$ divide $x-y$. This proves existence and uniqueness for two factors, and induction proves the general statement.

Nowand $12^3+1=1729$, so $x=12$ is a solution. For $1\le x<12$, one has $0<x^3+1<1729$, so no such positive integer can satisfy the congruence. Hence the smallest isModulo $7,13,19$, the roots are respectivelyEach list has three elements, and the Chinese remainder theorem combines the choices independently. Therefore the number of solutions with $1\le x\le 1729$ is

Let $V_n$ be the space of homogeneous degree-$n$ polynomials in $x,y$. For $g\in SU(2)$ and a column vector $v=(x,y{)}^T$, defineSubstitution preserves degree, andso ${\rho }_n:SU(2)\to \mathrm{GL}(V_n)$ is a continuous representation of a topological group. This is the homogeneous polynomial representation of SU2.

To prove irreducibility, restrict to the diagonal circleThe monomials $x^{n-k}{y}^k$ are its one-dimensional weight spaces, with distinct weights up to our harmless inverse-action convention. If $0\ne W\subseteq V_n$ is invariant, Fourier projection along this circle shows that $W$ contains a monomial. Differentiating the action of $SU(2)$ and complexifying gives the operatorsRepeated applications of $E$ and $F$ connect every monomial $x^{n-k}{y}^k$ to every other one. Hence $W$ contains the entire monomial basis, so $W=V_n$.

Every $g\in SU(2)$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$. Reading the eigenvalues on the monomial basis gives the character of the homogeneous polynomial representation of SU2For $z=e^{i\theta }$ this iswith the values at $\sin \theta =0$ obtained by continuity.

For completeness, let $W$ be any finite-dimensional irreducible continuous complex representation of $SU(2)$. Its restriction to the diagonal circle splits into integral weight spaces. Choose a vector of largest weight $n$. The raising operator kills it, and the ${\mathfrak{sl}}_2$ commutation relations show that $n$ is a nonnegative integer and that successive applications of the lowering operator form a string of weightsTheir span is an invariant copy of $V_n$; irreducibility forces it to equal $W$. Thus the classification of finite-dimensional representations of SU2 says

If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, then the eigenvalues on ${\bigwedge }^{2}V$ are ${\lambda }_i{\lambda }_j$ for $i<j$. Therefore the character of an exterior square isThe exterior square of an SU2 irreducible representation or the Clebsch-Gordan decomposition for SU2 with flip parity givesThe dimensions $7+3=10=\left(\genfrac{}{}{0ex}{}{5}{2}\right)$ check the result.

The third elementary symmetric polynomial in the ${\lambda }_i$, together with Newton identities, givesFor the five-dimensional representation $V_4$, the canonical dualityfinishes the decomposition. The weights $4,2,0,-2,-4$ sum to zero, so $\det V_4$ is trivial, and every $SU(2)$ irreducible is self-dual. Consequently

The compactness theorem says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One implication follows by taking the same model. Conversely, if $T$ had no model, then by the Godel completeness theorem it would prove a contradiction. A formal proof uses only finitely many assumptions, so some finite subset of $T$ would already have no model. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that if an $L$-theory $T$ has an infinite model, then it has models of arbitrarily large cardinality; more precisely, it has a model of cardinality at least $\kappa$ for every cardinal $\kappa$. Add new constants $c_{\alpha }$ for $\alpha <\kappa$ and the sentencesEvery finite subset of the enlarged theory can be interpreted in the given infinite model, since it mentions only finitely many constants. Compactness supplies a model of the whole enlarged theory, in which the $c_{\alpha }$ are pairwise distinct. Its reduct to $L$ is a model of $T$ having at least $\kappa$ elements. If $\kappa \ge |L|+{\aleph }_0$, the Downward Lowenheim-Skolem theorem gives a model of cardinality exactly $\kappa$.