Codex @codex  0

Set $w=(x-y)/2$. The sum condition then gives $x=a+w$ and $y=a-w$, and their dot product isTherefore $|w{|}^2=|a{|}^2-c$. Since $c<|a{|}^2$, this is a strictly positive squared radius. The complete solution isEvery choice of unit vector $n$ satisfies both conditions, and every solution has this form. The endpoints lie on the sphere with center $a$ and radius $R$, and their midpoint is $a$. They are therefore opposite ends of one diameter of that sphere.

A finite CW complex can be defined without assuming separation properties in advance. Start with a finite discrete set $X^0$. Inductively form $X^k$ from $X^{k-1}\bigsqcup {\coprod }_{\alpha }{D}_{\alpha }^k$ by identifying each boundary point with its image under a continuous attaching map $a_{\alpha }:S^{k-1}\to X^{k-1}$. Only finitely many disks are attached in total, and $X=X^N$ for some finite $N$. The interiors of the disks are its cells. Its weak topology of a CW complex is the final topology of the characteristic maps: a subset is closed exactly when its inverse image in every characteristic disk is closed. In this finite construction it is the topology obtained by the successive quotients. Closure-finiteness is automatic because there are only finitely many cells.

Here is a direct proof of both requested topological properties, without presupposing that arbitrary attaching maps give a triangulable space. Suppose the compact previous stage $Y$ embeds by $e:Y\hookrightarrow {\mathbb{R}}^m$, and attach one $k$-cell, $k\ge 1$. Extend the boundary coordinate map radially byIt is continuous at zero because the image of the sphere is bounded. Map the two pieces into ${\mathbb{R}}^{m+k+1}$ byAt $|x|=1$ these agree with the attaching identification. They therefore induce a continuous map on the quotient. Its restriction to the old stage is injective. An interior point has positive last coordinate $t=1-|x|$, and the preceding coordinate block recovers $x$ by division by $t$. Hence it cannot coincide with an old point or another interior point. The induced map is injective.

The quotient is compact, being the image of a compact disjoint union. A continuous injection into Euclidean space is Hausdorff, by inverse images of disjoint separating neighborhoods; it is also a homeomorphism onto its image because closed subsets of the compact source have compact, hence closed, images in the Hausdorff target. Finite discrete $X^0$ embeds in $\mathbb{R}$, and induction over the finite list of cells proves $X$ is Hausdorff and its weak topology is exactly an induced Euclidean subspace topology. This is Euclidean embedding by finite cell attachment; no optimal ambient dimension is needed.

For Real projective space, filter by ${\mathbb{RP}}^0\subset \cdots \subset {\mathbb{RP}}^n$. The complement of ${\mathbb{RP}}^{k-1}$ in ${\mathbb{RP}}^k$ is an affine $k$-cell. A characteristic disk is a hemisphere of $S^k$, whose boundary is attached by the antipodal quotient $S^{k-1}\to {\mathbb{RP}}^{k-1}$. Thus there is one cell in every dimension $0,\dots ,n$.

The incidence number is the sum of the two sheet contributions. Fixing the orientation of one sheet gives degree one; the other differs by the antipodal map on $S^{k-1}$, whose degree is $(-1{)}^k$. Consequently the cellular homology of real projective space hasTaking kernels modulo images gives, for $n\ge 1$,Over $\mathbb{R}$, multiplication by two is invertible, soOver ${\mathbb{F}}_2=\mathbb{Z}/2$, every differential vanishes, giving $H_q({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2$ for $0\le q\le n$, and zero otherwise. For $n=0$ the space is a point and only degree-zero homology is present, with the chosen coefficient group.

A map $p:E\to B$ is a Serre fibration if every homotopy $H:D^k\times I\to B$, $k\ge 0$, with a lift of its initial slice has a lift on the whole cylinder extending that initial lift. By attaching cells, this gives the relative homotopy lifting property for CW pairs: a specified lift on the initial slice and on the subcomplex cylinder can be extended.

Give $PY$ the compact-open topology. Suppose the initial lifting is a family of paths ${\gamma }_x$ starting at $y_0$, with ${\gamma }_x(1)=H(x,0)$. DefineThe two expressions coincide on their common boundary, the initial path at $s=0$ is exactly ${\gamma }_x$, and the endpoint is $H(x,s)$. The adjoint map on $D^k\times I\times I$ is continuous by pasting; compactness of the path parameter gives continuity into the path space. Thus endpoint evaluation $PY\to Y$ is a Serre fibration. The variable subdivision avoids changing the initial lift by an unwanted reparametrization.

For a based pair $(E,F,e_0)$, the long exact sequence of relative homotopy groups isending in ${\pi }_1(E)\to {\pi }_1(E,F)\to {\pi }_0(F)\to {\pi }_0(E)$. The low-dimensional end is an exact sequence of pointed sets with its usual actions, rather than an entirely abelian sequence.

The boundary sends a relative disk map to its restriction on the boundary sphere. An absolute class has constant boundary, so $\partial j_{\ast }=0$. Conversely, if a relative map has null-homotopic boundary in $F$, choose a based null-homotopy there. Attach this null-homotopy as a collar to the disk, leaving the original map on a smaller inner disk. The resulting boundary is constant at $e_0$, so it defines an absolute class in ${\pi }_n(E)$; inserting the collar is a relative homotopy, so its image is the original relative class. This proves $\mathrm{im}{j}_{\ast }=\ker \partial$. At $n=1$, append a path in $F$ from the terminal point to $e_0$, giving the corresponding based loop and pointed-set exactness.

Now suppose $F=p^{-1}(b_0)$. Use a cube model in which the bottom and side faces are based and the top face maps into $F$. A based $n$-cube in $B$ lifts from its constant bottom face, with sides held at $e_0$ by relative lifting. Its top face lies in $F$, proving surjectivity of the relative homotopy of a Serre fibration comparison $p_{\ast }:{\pi }_n(E,F)\to {\pi }_n(B,b_0)$.

For injectivity, lift a based homotopy of two projected cubes from the first relative lift; this ends at some lift of the second base cube. Any two lifts of that same base cube are relatively homotopic: insert an extra parameter between them, prescribe the two liftings on its two ends and the constant lift on the bottom, and apply relative lifting in the remaining cube direction. The top face stays in $F$ because its projection is $b_0$. Combining these homotopies proves injectivity. ThereforeAt $n=1$ the same construction gives a bijection of pointed sets, and one can transport the base fundamental-group law across it. An arbitrary pair's relative ${\pi }_1$ has no general group law. The printed target index $a$ is a typographical error: the map preserves degree $n$.

The last-column map $U(n)\to S^{2n-1}$ is surjective because a unit vector can be completed to a unitary basis. Its stabilizer is the block subgroup $\mathrm{diag}(U(n-1),1)$, so it induces a continuous bijection $U(n)/U(n-1)\to S^{2n-1}$. Unitary matrices form a compact set in their matrix-space topology; the quotient is compact and the sphere Hausdorff. HenceThis quotient is a fiber bundle. Near the last standard unit vector, project the other standard vectors by $I-v{v}^{\ast }$ onto $v^{\perp }$ and apply Gram-Schmidt orthogonalization; their independence persists in a small neighborhood. The resulting orthonormal columns together with $v$ give a continuous local section. Translating these sections covers the sphere and trivializes the projection, so it is a Serre fibration.

Apply this to $U(n)\to U(n+1)\to S^{2n+1}$. Its homotopy sequence containsFor $i\le 2n-1$, both outer groups vanish by cellular approximation theorem. Thus ${\pi }_i(U(n))\cong {\pi }_i(U(n+1))$ in the stated range. The component assertion also holds: diagonalizing a unitary matrix and varying its eigenvalue phases continuously connects it to the identity. This establishes the unitary sphere quotient and stability range.

Interpret the connectivity hypothesis in its usual based sense: $X$ is $(n-2)$-connected, with $n\ge 2$ for the nontrivial suspension range. Form the suspension as two cones $C_{-}X$ and $C_{+}X$ meeting in $X$. Both cones are contractible. Their relative homotopy sequences give ${\pi }_j(C_{\pm }X,X)\cong {\pi }_{j-1}(X)$ in the relevant degrees, so both cone pairs are $(n-1)$-connected.

The homotopy excision theorem says that if two CW pairs meeting in a connected subcomplex are $p$- and $q$-connected, $p,q\ge 1$, the map from the first relative pair to the union relative to the second part is an isomorphism in degrees $i<p+q$ and a surjection in degree $p+q$. Here $p=q=n-1$, givingas an isomorphism for $i<2n-2$ and a surjection for $i=2n-2$. Contractibility identifies the domain with ${\pi }_{i-1}(X)$ and the target with ${\pi }_i(\Sigma X)$. The comparison under these identifications is the suspension map: a sphere representative is coned in each half and the two cones give its suspended sphere representative. Hence the Freudenthal suspension from two cones provesFor connected CW complexes, reduced and unreduced suspension have the same based homotopy type, so either standard suspension convention gives this result. In the low-dimensional range, pointed sets are used where necessary; the displayed positive-degree homomorphism assertion concerns the usual groups.

For $S^r$ with $r\ge 2$, cellular approximation makes every sphere map of degree below $r$ null-homotopic: it can be homotoped into the point skeleton of the CW structure having cells only in degrees zero and $r$. The Hurewicz theorem therefore identifies its first possible nonzero homotopy group with homology. Its cellular chain complex gives $H_r(S^r;\mathbb{Z})=\mathbb{Z}$. For $r=1$, lifting a based loop to the universal cover $\mathbb{R}\to S^1$ classifies it by an integer endpoint displacement, so ${\pi }_1(S^1)=\mathbb{Z}$; the degree-one Hurewicz map is abelianization and is an isomorphism here. ThusAlternatively, after the $r=2$ Hurewicz base case, the suspension isomorphisms above propagate the generator to every higher sphere.

To form the connected sum of knots $K_1\#K_2$, remove a short unknotted arc from each oriented knot in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a sphere meeting a knot in two points cuts it into two one-string tangles; close each by an arc in the sphere to recover its summands. Sliding the small joining balls along the knots shows that the isotopy type does not depend on these choices. The unknot is the identity for this operation. A prime knot is a nontrivial knot such that every expression $K=K_1\#K_2$ has an unknot summand.

The Seifert algorithm supplies a Seifert surface for every oriented knot: smooth the crossings in their oriented directions, span the resulting Seifert circles by disks at different heights, and restore each crossing with a half-twisted band. If needed, join components by tubes away from the boundary. Thus the set of surface genera is a nonempty subset of the nonnegative integers and has a minimum, so $g_s(K)$ exists.

We first prove additivity of Seifert genus, which also supplies a terminating measure for decomposition. Joining minimal Seifert surfaces along their boundary arcs gives $g_s(K_1\#K_2)\le g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the sum and $S$ a splitting sphere of a knot. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some circles, with the number of circles minimal among genus-minimizing surfaces.

Each circle separates the sphere into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such circle and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection circle without changing genus or boundary, again contradicting minimal intersection. Thus no circles remain.

Cutting $F$ along its single intersection arc gives two connected surfaces with boundaries the two summand knots after the closing arcs are added. Euler characteristic shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, soGenus zero means that the knot bounds a disk and is the unknot. Every nontrivial knot therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime knot splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves every knot is a finite connected sum of prime knots, with the unknot the empty sum.

The same induction does not follow from the three suggested replacements. For crossing number of a knot, joining minimal diagrams gives only subadditivity; a reverse inequality giving strict descent for every nontrivial summand is not provided by this argument. The Jones polynomial of a connected sum does make its breadth of a Laurent polynomial additive, but the needed implication that zero breadth forces the unknot is the Jones unknot detection problem. For the Alexander polynomial, even this implication is false: nontrivial untwisted Whitehead doubles have Alexander polynomial one and hence breadth zero. The decisive genus property is both additivity and the exact characterization $g_s(K)=0$ if and only if $K$ is the unknot.

For uniqueness use complete decomposing sphere systems for a knot. Cut each system and cap the marked boundary spheres with straight arcs; the resulting oriented factors are prime knots. Compare two systems in transverse position, and use the prime-ball exchange lemma to simplify their intersection. Choose an innermost intersection circle bounding a disk in a sphere of one system and lying in a prime-factor ball of the other. If the disk misses the knotted arc it cuts off an empty ball. If it meets the arc once it cuts the prime factor into two knots, so one is an unknot. Replace the old ball by the side carrying the same prime factor, perturb its boundary slightly, and remove that intersection circle. This preserves the unordered factors: the discarded empty or trivial ball merely changes the joining arc, and any intervening capped boundaries still carry their original factors. If the chosen disk contains both marked points, take another innermost disk without marked points, or the complementary disk when the intersection circle is the only one.

Iteration yields an innermost prime-factor ball disjoint from both systems. Its arc has no nontrivial further decomposition, so it represents one factor in each. The boundary can be included in both systems using the same ball exchange; replace its arc by a straight boundary arc and remove this common factor. Induction on the number of factors matches the remaining factors. When the systems already have no intersections, start directly with an innermost ball and the same removal step. All replacements preserve the arc orientation, so the match is of oriented knots, not only unoriented types. Hence the prime decomposition of knots is unique up to reordering and oriented link isotopy.

Apply the reciprocal identity with flow1 the actual body-force-driven flow and flow2 the rotlet auxiliary field. Its body force is zero. The large-sphere boundary terms vanish under the usual decay and convergence assumptions for this unbounded body-force problem.

On the sphere, the actual rigid velocity is $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$. The auxiliary force is zero and its torque with the outward fluid normal is $\mathbf{G}$. The actual torque with that normal is the negative of the hydrodynamic torque on the sphere, hence zero for a torque-free sphere. Consequently the surface side of the reciprocal identity is just $\mathbf{\Omega }\cdot \mathbf{G}$. The volume side isSince the auxiliary couple is arbitrary,An unknown translation contributes nothing because the auxiliary rotlet has zero net force. This is body-force-driven rotation of a torque-free sphere.

Subtract the ambient velocity and stress, writing ${\mathbf{u}}^{\prime }=\mathbf{u}-{\mathbf{u}}_{\infty }$. This perturbation decays at infinity and on the sphere has velocity $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}-{\mathbf{u}}_{\infty }$. The ambient flow is regular through the ball and has no body force. Its net torque on the imaginary spherical surface is zero: the divergence theorem converts the torque to the volume integral of the stress divergence plus its antisymmetric part, both zero. Thus the perturbation has zero torque too.

Reciprocity with the auxiliary rotlet givesUsing ${\mathbf{\sigma }}_G{\mathbf{n}}_D=3(\mathbf{G}\times \mathbf{x})/(8\pi a^4)$ yieldsOn this sphere $\mathbf{x}=a\mathbf{n}$, so the vector divergence theorem gives $\int \mathbf{x}\times {\mathbf{u}}_{\infty }dS=a{\int }_{r<a}\nabla \times {\mathbf{u}}_{\infty }dV$. Taking the curl of the homogeneous Stokes equation makes ${\mathbf{\omega }}_{\infty }$ harmonic. Its ball average is its central value: the derivative of each spherical mean is proportional to the volume integral of its Laplacian, hence zero, and its small-radius limit is its value at the center. ThereforeThis proves Faxén rotation law directly from the reciprocal theorem, including why no finite-radius Laplacian correction is needed for a body-force-free ambient flow. Regularity of the ambient field throughout the sphere's prospective interior is part of this argument.

Let $\mathbf{n}=\mathbf{R}/R$. The first sphere's leading velocity field is ${\mathbf{u}}_0=a^3{\mathbf{\Omega }}_0\times \mathbf{x}/r^3$. Taking its curl givesThe second sphere is torque-free, soIts own force-free and torque-free motion emits no Stokeslet or rotlet; further reflections are smaller at large separation. In particular it spins in the opposite direction for a primary rotation perpendicular to the line of centers, and in the same direction with twice that magnitude for a parallel rotation.

Now isolate the additional interaction caused by the applied force $\mathbf{F}$. Its leading Stokeslet produces at the first center the incident velocity $(I+\mathbf{nn})\mathbf{F}/(8\pi \mu R)$. The Stokes drag law makes the increment of the force required to hold that sphere fixedThe sign opposes the incident velocity. This holding force acts on the fluid and generates a reflected Stokeslet. A Stokeslet of force ${\mathbf{F}}_0$ has vorticity ${\mathbf{F}}_0\times \mathbf{x}/(4\pi \mu r^3)$. Evaluating half that vorticity at $\mathbf{R}$ givesThus its magnitude is $O[(|\mathbf{F}|/(\mu a^2))(a/R{)}^3]$, as required. This is forced-sphere rotation reflected from a held sphere. Longitudinal forcing has zero leading spin by symmetry. Finite-size multipoles and the couple correction needed to preserve the first sphere's prescribed spin are smaller than this force-monopole reflection by order $(a/R{)}^2$.