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Over an algebraically closed field of characteristic different from two, a projective change of coordinates puts every smooth quadric surface in the formThe Segre embedding identifies this surface with ${\mathbb{P}}^1\times {\mathbb{P}}^1$ byTwo distinct fibres of either coordinate projection are disjoint lines in one of the rulings of a smooth quadric surface. Explicitly,both lie on $V$ and have no common projective point. Thus $V$ contains a pair of disjoint lines.

Because $k$ is algebraically closed, the smooth quadric surface $V$ has a $k$-point $p$. Choose a plane $\Pi \cong {\mathbb{P}}_k^2$ not containing $p$ and project from $p$:For a general point $q\in \Pi$, the line $\overline{pq}$ meets the degree-two surface $V$ in $p$ and exactly one further point. Sending $q$ to that second intersection is a rational map inverse to ${\pi }_p$ on dense open subsets. Therefore

For $x\in X$, define $e_x\in X^{\ast \ast }$ by $e_x(\varphi )=\varphi (x)$. The map $J:x\mapsto e_x$ is linear, andshows $\Vert Jx\Vert \le \Vert x\Vert$, so $J$ is continuous. The granted equality $\Vert Jx\Vert =\Vert x\Vert$ is the canonical embedding into the bidual.

Suppose, contrary to the claim, that exactly the same linear functionals are continuous for $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$. Their common continuous dual has two operator norms, say $\Vert \cdot {\Vert }_{\ast }$ and $\Vert \cdot {\Vert }_{\ast }^{\prime }$. Both dual spaces are Banach, even if the second normed space $X$ is not complete. The identity maphas a closed graph: convergence in either operator norm implies pointwise convergence on $X$, so two limits must agree. The closed graph theorem makes the two dual norms equivalent.

Applying the isometric bidual formula for each primal norm then gives constants $c,C>0$ such thatThis says the two norms are equivalent norms, contrary to the hypothesis. Their continuous duals must therefore differ as sets, so a linear functional belonging to one and not the other is continuous for exactly one of the two norms.

For integers $m,n\ge 1$, let $E_{m,n}$ contain those $f\in C([0,1])$ for which there is some $x\in [0,1]$ such thatwhenever $|y-x|<1/n$. Compactness of $[0,1]$ shows that $E_{m,n}$ is closed in the uniform norm: for a uniformly convergent sequence, choose convergent subsequences of the corresponding witness points and pass to the limit.

Each $E_{m,n}$ has empty interior. Given $f$ and a uniform-error tolerance, first approximate $f$ by a polygonal function, then add a sufficiently small-amplitude, sufficiently high-frequency sawtooth whose slopes dominate both $m$ and every slope of the polygonal approximation. At every point, one can move a distance below $1/n$ along one adjacent linear segment and obtain a difference quotient of magnitude greater than $m$. The perturbed function lies in the prescribed ball but outside $E_{m,n}$.

If $f$ is differentiable at some point, its nearby difference quotients are bounded, so $f\in E_{m,n}$ for some $m,n$. The functions differentiable somewhere therefore form a meagre set. By the Baire category theorem, their complement is nonempty; indeed it is dense. Thus

The Baire category theorem says that a complete metric space cannot be a countable union of nowhere dense sets, equivalently that every countable intersection of open dense sets is dense. To prove it, let $G_1,G_2,\dots$ be open dense subsets and let $V$ be nonempty and open. Choose nested closed ballsstarting inside $V\cap G_1$. The centres are a Cauchy sequence; completeness gives a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense.

Choose a countable sequence $q_j\uparrow p$ with $1\le q_j<p$. Since ${\ell }^q\subset {\ell }^r$ for $q<r$, every ${\ell }^q$ with $q<p$ lies in some ${\ell }^{q_j}$. For integers $m\ge 1$, setPointwise convergence and Fatou lemma for series show that $E_{j,m}$ is closed in ${\ell }^p$. It has empty interior: inside any ${\ell }^p$ ball, add a sufficiently long finite block whose ${\ell }^p$ norm is tiny but whose ${\ell }^{q_j}$ norm is larger than $m$. Therefore every $E_{j,m}$ is nowhere dense. Since ${\ell }^p$ is a Banach space, Baire gives

An explicit element of the difference isIts $p$th powers form a convergent logarithmic series, whereas for every $q<p$ the powers contain $n^{-q/p}$ with exponent below one and their series diverges. Hence $x\in {\ell }^p$ but $x\notin {\ell }^q$ for every $q<p$.

Let $F_2=⟨a,b⟩={\pi }_1(S^1\vee S^1)$ and take the connected two-sheeted covering graph corresponding to the kernel ofIt has two vertices, a $b$-loop at each vertex, and two lifted $a$-edges joining the vertices. Choose one lifted $a$-edge as a spanning tree. The other three edges give the based loopsCollapsing the spanning tree identifies the cover with a wedge of three circles, so these loops freely generate its fundamental group. Hence they generate an explicit subgroupisomorphic to $F_3$, in agreement with the Reidemeister–Schreier theorem.

Cover the attached disc by a slightly enlarged interior and a collar of its boundary, and apply the Seifert-van Kampen theorem. The collar intersection has fundamental group generated by the boundary loop, whose image in $X$ is $[\alpha ]$, while that loop becomes trivial in the disc. Van Kampen therefore quotients by the smallest normal subgroup containing $[\alpha ]$:This is the general fundamental group after attaching a 2-cell.

Start with a wedge of two circles, whose fundamental group is the free groupAttach a disc along the loop represented by $a^3{b}^{-7}$. The displayed quotient gives a connected two-dimensional CW complex $X$ with

Let $i:Y\hookrightarrow M_f$ be the canonical inclusion and defineThis is well-defined because $[0,x]=f(x)$ in the mapping cylinder, and continuity follows from the universal property of the quotient topology. Clearly $r\circ i={\mathrm{id}}_Y$.

DefineThe formula respects the gluing relation at $t=0$, so it descends from the disjoint union to a continuous homotopy. It satisfies $H_0={\mathrm{id}}_{M_f}$ and $H_1=i\circ r$. Thusindeed $Y$ is a deformation retract of $M_f$.

Let ${\sigma }_1,\dots ,{\sigma }_n:K\hookrightarrow \mathbb{C}$ be the embeddings of the number field $K$. The discriminant of elements of a number field isIf the ${\alpha }_j$ are $\mathbb{Q}$-linearly dependent, the embedding matrix has dependent columns and the discriminant vanishes. Conversely, if they are independent, they form a $\mathbb{Q}$-basis of the $n$-dimensional vector space $K$. The trace pairing of a characteristic-zero number field is nondegenerate, so its Gram matrix in this basis is nonsingular. Hence

Let $x_1,\dots ,x_5$ be algebraically independent indeterminates and set $a_i=e_i(x_1,\dots ,x_5)$, the elementary symmetric polynomials. ThenThe Fundamental theorem of symmetric polynomials identifies the fixed field of the natural $S_5$ action on $\mathbb{Q}(x_1,\dots ,x_5)$ with $\mathbb{Q}(a_1,\dots ,a_5)=K$. Every permutation of the roots therefore gives a $K$-automorphism, and the Galois group of a polynomial isA characteristic-zero polynomial is solvable by radicals exactly when its Galois group is a solvable group. Part (c) shows that $S_5$ is not solvable, so

Let $\sigma (\alpha )={\xi }_d\alpha$ with $\alpha \ne 0$. Thenand these $d$ elements are distinct because ${\xi }_d$ is a primitive root of unity. Thus the stabilizer of $\alpha$ in the cyclic Galois group is trivial. The orbit-stabilizer form of the Fundamental theorem of Galois theory givessoMoreover,The fixed field of the whole Galois group is $K$, and thereforeThis is the basic eigenvector mechanism behind a cyclic Kummer extension.

Every $\tau \in \mathrm{Aut}(L({\xi }_N)/K({\xi }_N))$ fixes $K$ and permutes the roots of $f$, so it preserves their splitting field $L$. Restriction therefore defines a homomorphismIf $\rho (\tau )$ is the identity, then $\tau$ fixes both $L$ and ${\xi }_N$, which generate $L({\xi }_N)$; hence $\tau$ is the identity. Thus $\rho$ is injective, and its image is a subgroup:

A 3-connected graph contains a cycle, and any cycle is a complete graph subdivision $T{K}_3$.

For $T{K}_4$, choose a cycle $C$ for which some $C$-bridge has at least three attachment vertices $a,b,c$; such a choice is always possible in a 3-connected graph. Indeed, every component outside a cycle has at least three attachments, since one or two would form a vertex cut, and if a chosen cycle is spanning, a chord can be used to choose a smaller cycle and expose the remaining bridge. In a minimal connected part of that bridge joining $a,b,c$, suppress degree-two vertices. The result has a branch vertex $w$ joined to $a,b,c$ by three paths whose interiors are mutually disjoint and avoid $C$. The three arcs of $C$ between $a,b,c$, together with these three paths, form a subdivision of the complete graph on branch vertices $w,a,b,c$. Hence

A 3-connected graph need not contain a $T{K}_5$. The complete graph $K_4$ itself is 3-connected and has only four vertices, so it cannot contain a graph subdivision of $K_5$. Equally, any planar 3-connected graph is a counterexample because a subdivision of the complete graph $K_5$ would be nonplanar.

For nonadjacent vertices $a,b$, the local connectivity $\kappa (a,b;G)$ is the minimum cardinality of an $a$-$b$ vertex separator: a set $S\subseteq V(G)\setminus \{a,b\}$ such that $a$ and $b$ lie in different components of $G-S$.

To prove the vertex form of the Menger theorem, replace every vertex $v\notin \{a,b\}$ by an arc $v_{\mathrm{in}}\to v_{\mathrm{out}}$ of capacity one. Replace every undirected edge $uv$ by directed arcs from $u_{\mathrm{out}}$ to $v_{\mathrm{in}}$ and conversely, each of capacity larger than $|V(G)|$; leave $a,b$ unsplit. An integral $a$-$b$ flow of value $k$ decomposes into $k$ paths, and the unit-capacity vertex arcs force the corresponding graph paths to be internally vertex-disjoint paths. Conversely, $k$ such paths give an integral flow of value $k$.

Because $a$ and $b$ are nonadjacent, any finite minimum cut consists entirely of unit-capacity vertex arcs. These arcs correspond exactly to an $a$-$b$ vertex separator of the same size. The max-flow min-cut theorem, whose augmenting-path proof also guarantees an integral maximum flow for integral capacities, therefore givesIn particular, $G$ contains the required set of $\kappa (a,b;G)$ paths.