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Write the diagonal maximal torus of $U(n)$ as $t=\mathrm{diag}(z_1,\dots ,z_n)$, $|z_i|=1$, and let $\delta =(n-1,n-2,\dots ,0)$. Irreducible complex representations are indexed by dominant integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$, their highest weights. The Weyl character formula in integer-exponent form isThe ratio initially uses distinct $z_i$, and extends to repeated eigenvalues by continuity. Every unitary matrix is conjugate to such a diagonal matrix, so this determines the character on the entire group. The shift $\delta$ differs from the usual Weyl vector $\rho =((n-1)/2,(n-3)/2,\dots ,(1-n)/2)$ by a multiple of $(1,\dots ,1)$; the corresponding determinant factors cancel between numerator and denominator.

We first justify the highest-weight information needed in the proof. Restriction to the torus splits a representation into integer weight spaces. In an irreducible representation choose a lexicographically largest weight $\lambda$ and a nonzero vector $v$ of that weight. A matrix unit $E_{ij}$ with $i<j$ raises the weight by $e_i-e_j$, so it annihilates $v$. The character is invariant under permutations of diagonal coordinates, because permutation matrices conjugate the torus. Hence permutations of $\lambda$ are also weights, and lexical maximality forces ${\lambda }_1\ge \cdots \ge {\lambda }_n$.

Irreducibility implies that $v$ generates the representation under the complexified Lie algebra. Indeed, a Lie-algebra invariant subspace is preserved by exponentials of skew-Hermitian matrices, which generate $U(n)$. Reorder any product of matrix units into lower-triangular units, diagonal units and upper-triangular units, usingInduction on length and on the number of out-of-order pairs gives the required spanning version of the Poincaré-Birkhoff-Witt theorem. Upper-triangular units kill $v$, and diagonal units act by scalars. Every nonempty product of lower-triangular units lowers its weight by a nonzero sum of positive roots. Therefore the top weight space is exactly $\mathbb{C}v$, and its multiplicity is one. This supplies the leading coefficient without assuming a character formula.

For completeness, every dominant integer tuple occurs. Put $d_j={\lambda }_j-{\lambda }_{j+1}\ge 0$ and form the representationIts lexicographically highest weight is $\lambda$, with one-dimensional highest space, represented by the tensor products of $e_1\wedge \cdots \wedge e_j$. A negative determinant exponent is a legitimate one-dimensional unitary-group character. Averaging an inner product over normalized Haar measure gives invariant orthogonal complements, so the tensor representation splits into irreducible summands. One of them has highest weight $\lambda$. Uniqueness will also follow from the formula and Schur orthogonality relations.

Define the alternant $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ for a strictly decreasing integer tuple $\kappa$. The denominatoris the Vandermonde determinant, and its modulus is the denominator modulus in the Weyl integration formula. The Laurent polynomial ${\chi }_{\lambda }{A}_{\delta }$ is alternating. Any alternating Laurent polynomial decomposes as a finite summonomials with repeated exponents have zero coefficient, and the remaining permutation orbits are indexed uniquely by strictly decreasing tuples.

The lexicographically highest monomial in the product is $z^{\lambda +\delta }$ with coefficient one. Indeed, the highest monomial of the character is $z^{\lambda }$ with coefficient one, and the unique lexical maximum among permutations of $\delta$ is $\delta$ itself. It follows that $c_{\lambda +\delta }=1$.

Fourier orthogonality on the torus, whose probability measure is ${\prod }_{i}d{\theta }_i/(2\pi )$, gives alternant orthogonality on a unitary torus:Expand both determinants: an integral survives exactly when the two exponent tuples are identical up to permutation. Strict decreasing order then forces $\kappa =\eta$, and its $n!$ matching terms each contribute one.

Finally, Schur orthogonality relations give ${\int }_{U(n)}|{\chi }_{\lambda }{|}^2=1$. Applying the allowed Weyl integration formula to this class function givesSince the coefficient $c_{\lambda +\delta }$ already equals one, every other coefficient vanishes. Thus ${\chi }_{\lambda }{A}_{\delta }=A_{\lambda +\delta }$, proving the formula. If two irreducible representations had the same highest weight, their characters would now coincide; the Schur orthogonality relations imply that they are isomorphic. Apparent singularities at repeated diagonal eigenvalues are removable because the quotient equals the actual finite Laurent-polynomial character. No dimension formula has been assumed.

Let $V={\mathbb{C}}^n$ and $m\ge 0$. Identify the symmetric power ${\mathrm{Sym}}^{m}V$ with the degree-$m$ part of the symmetric algebra on the basis vectors $e_1,\dots ,e_n$, using formal variables $x_i$ for these vectors. Its monomial basis isThe diagonal maximal torus of the unitary group acts on this monomial by the character $z_1^{a_1}\cdots z_n^{a_n}$. Distinct tuples give distinct characters, so each weight space is one-dimensional.

Suppose $M$ is a nonzero invariant complex subspace. For a nonzero vector $w\in M$, select a monomial with nonzero coefficient and extract it by torus averaging:Fourier orthogonality makes this a nonzero scalar multiple of $x^a$. It belongs to $M$, so $M$ contains a monomial.

Differentiating the unitary group action makes $M$ invariant under $\mathfrak{u}(n)$ and hence under its complexification ${\mathfrak{gl}}_n(\mathbb{C})$. The matrix unit $E_{ij}$ replaces an occurrence of $e_j$ by $e_i$, so its action on the symmetric algebra isWhenever $a_j>0$, this transfers one unit of degree from coordinate $j$ to coordinate $i$ with a nonzero coefficient. Starting with any monomial, such transfers can produce every other tuple of nonnegative integers summing to $m$. Thus $M$ contains every monomial and equals ${\mathrm{Sym}}^{m}V$.

ConsequentlyFor $m=0$ this is the one-dimensional trivial representation, and for $n=1$ every symmetric power is also one-dimensional. This weight-transfer proof of irreducibility of symmetric powers uses neither the Weyl character formula nor the Weyl dimension formula, so no unproved dimension formula enters the argument.

The displayed torus-only Weyl integration formula needs $\phi$ to be a class function, meaning $\phi (xg{x}^{-1})=\phi (g)$. This hypothesis is implicit in the usual formula but is not printed here. We prove the general version for a continuous function, then specialize:Here $dg$ and $dt$ are probability Haar measures; $d\mu$ is the invariant probability measure on the homogeneous space $G/T$; and $W=N_G(T)/T$ is the finite Weyl group.

We use the following standard maximal-torus facts, as permitted: every element of a connected compact Lie group lies in some maximal torus; all maximal tori are conjugate; two elements of a fixed $T$ are conjugate in $G$ exactly when they are in the same $W$-orbit; and the complexified Lie algebra decomposes into ${\mathfrak{t}}_{\mathbb{C}}$ and one-dimensional root spaces. Opposite roots of a root system give real two-dimensional planes in ${\mathfrak{t}}^{\perp }$. At an element $t$ for which no root character has value one, the identity component of its centralizer is $T$.

Choose a positive root system ${\Phi }^{+}$. Regard each root $\alpha$ as a character $\alpha :T\to U(1)$ through the Adjoint representation. A convenient globally defined denominator isThe usual Weyl denominator additionally has a unit-modulus factor $e^{\rho }(t)$, with $\rho$ half the sum of positive roots. Where that half-weight is not a global torus character, it may be interpreted on a covering torus; the squared modulus is nevertheless unambiguous. In either convention,

Average an inner product on $\mathfrak{g}$ over the Adjoint representation to make it invariant. It defines a bi-invariant Riemannian metric on $G$ and the quotient metric on $G/T$. Consider the conjugation mapAt $(T,t)$, identify the source tangent directions with $X\in {\mathfrak{t}}^{\perp }$ and $Y\in \mathfrak{t}$, and translate the target tangent vector to the identity by $t^{-1}$. Differentiating givesThe two summands are orthogonal. On a real root plane, $\mathrm{Ad}(t^{-1})$ is a rotation with complex eigenvalues $\alpha (t{)}^{-1},\alpha (t)$. Thus the absolute real Jacobian determinant on this plane is $|1-\alpha (t){|}^2$. The torus direction contributes one. Therefore the conjugation Jacobian for a compact Lie group isThis is also the Jacobian for the normalized measures: the Riemannian submersion has fibers isometric to $T$, so $\mathrm{vol}(G)=\mathrm{vol}(G/T)\mathrm{vol}(T)$; the probability-volume normalization factors cancel.

Let $T_{\mathrm{reg}}$ be the elements for which $J_q\ne 0$. The restriction of $q$ is a local diffeomorphism onto the regular elements of $G$. It is $|W|$-to-one. Indeed, fixing $t_0\in T_{\mathrm{reg}}$, any pair with $xt{x}^{-1}=t_0$ has $xT{x}^{-1}=T$, because conjugation identifies the identity components of the two centralizers. Thus $x\in N_G(T)$, and there is exactly one pair $(xT,x^{-1}{t}_{0}x)$ for each coset in $N_G(T)/T$. This argument does not require the full centralizer of $t_0$ to be connected.

The nonregular locus has measure zero. In $T$ it is a finite union of sets $\alpha (t)=1$ of smaller dimension; their conjugation images are images of smaller-dimensional manifolds under $q$, and hence have zero volume in $G$. The change-of-variables formula for the regular covering therefore givesFor a class function, the inner integral is $\phi (t)$, proving the required formula with its correct hypothesis.

Without conjugation invariance, the torus-only expression can be false. For $G=SU(2)$ with diagonal $T$, take $\phi (g)=|g_{12}{|}^2$. It vanishes on $T$, whereas its group integral is $1/2$: the two squared coordinates of a Haar-distributed unit column have equal means and sum to one. Thus the conjugacy average cannot simply be dropped.

Work in the based CW homotopy category. A homotopy-invariant functor depends on a map only through its based homotopy class. In a sequence of pointed sets, exactness at the middle means that the preceding image equals the inverse image of the distinguished point; it does not imply injectivity or surjectivity at the ends. For groups this is the familiar image-equals-kernel condition. A half-exact functor preserves such middle exactness on the appropriate elementary sequence.

For a contravariant functor $F$, the relevant elementary sequence is a cofibration $A\hookrightarrow X\to X/A$. Half-exactness isexact at $F(X)$. One should not silently turn this into a short exact sequence. In particular, arbitrary group actions and pointed sets at low degrees are not interchangeable with abelian groups.

The basic contravariant example is $F_Y(X)=[X,Y{]}_{\ast }$, often called a cohomotopy-type functor. Suppose a map $f:X\to Y$ becomes null-homotopic on $A$. A chosen null-homotopy extends it over the cone attached to $A$, giving a map from the topological mapping cone to $Y$. Conversely every such extension restricts to a null-homotopic map on $A$. Since the topological mapping cone of a subcomplex inclusion models $X/A$, this proves half-exactness, rather than merely asserting it.

Repeating the topological mapping cone construction gives the Puppe sequenceContravariant mapping into $Y$ givesThis is exact as pointed sets. An H-group structure on $Y$, for example a loop space with path concatenation, provides group structures; repeated looping gives abelian group structures in the usual ranges. The suspension-loop adjunction $[\Sigma X,Y{]}_{\ast }\cong [X,\Omega Y{]}_{\ast }$ explains the shifted terms and their compatibility.

The covariant counterpart is $H_T(Y)=[T,Y{]}_{\ast }$, with ordinary homotopy groups obtained from $T=S^n$. Its natural elementary sequence is a homotopy fiber sequence $F\to E\to B$, not an arbitrary cofibration. A map $a:T\to E$ whose composite to $B$ is null-homotopic, together with that null-homotopy, determines a map into the homotopy fibre. Conversely a fibre map projects to such a null composite. This proves exactness of $[T,F{]}_{\ast }\to [T,E{]}_{\ast }\to [T,B{]}_{\ast }$ at the middle. Adding the loop terms produces the homotopy long exact sequence. For a Serre fibration, its ordinary fibre supplies the same homotopy information.

This variance distinction matters. Include one circle, generated by $a$, in $S^1\vee S^1$, with other generator $b$. The quotient is the second circle. The kernel of ${\pi }_1(S^1\vee S^1)\to {\pi }_1(S^1)$ is the normal closure of $a$, which contains $ba{b}^{-1}$. The image of the included circle is only the cyclic subgroup generated by $a$, and does not contain that reduced word. Thus covariant ${\pi }_1$ is not half-exact on this cofibration. Fibre and cofibre statements cannot be swapped just by reversing notation.

Reduced cohomology gives group-valued contravariant examples. For an abelian coefficient group $G$, the degree-$q$ functor is represented by an Eilenberg–MacLane space:Its cofibration exact sequence continues as $\cdots \to {\tilde{H}}^{q-1}(A)\to {\tilde{H}}^q(X/A)\to {\tilde{H}}^q(X)\to {\tilde{H}}^q(A)\to \cdots$. Ordinary homotopy groups are instead covariant, with decreasing group degree across a fibre connecting map. These are the two complementary representable patterns, maps out of a test object and maps into a representing object.

Brown's representability theorem makes the contravariant pattern structural. A reduced contravariant homotopy functor on based CW complexes is represented by a CW complex if it converts arbitrary wedges to products and satisfies the Mayer-Vietoris weak-pullback gluing axiom. The latter says that compatible classes on two subcomplexes extend to their union. It contains more information than the single kernel statement of half-exactness. Consequently half-exactness alone is not a representability criterion. In a representing functor these axioms are natural: a map on a wedge is a family of based maps, while compatible maps on an intersection can be adjusted by a homotopy and glued across a homotopy pushout.

A reduced generalized cohomology theory with the usual wedge axiom supplies these axioms degree by degree. Its representing spaces $E_q$ have $E_q\simeq \Omega E_{q+1}$ by suspension compatibility, assembling an Omega-spectrum. Thus generalized cohomology can be expressed as $E^q(X)=[{\Sigma }^{\infty }X,{\Sigma }^{q}E{]}_{\mathrm{st}}$, while represented homology is $E_q(X)={\pi }_q(E\wedge {\Sigma }^{\infty }X)$. The appearance of mapping into a spectrum in one case and smashing with it in the other is another form of the same variance distinction.

There is a literal duality after stabilization and a finiteness restriction. Here a finite based CW complex is identified with its suspension spectrum. For a finite spectrum $X$, let $DX=F(X,\mathbb{S})$ be its Spanier-Whitehead dual. Then $D^{2}X\simeq X$ and $[X,Y{]}_{\mathrm{st}}\cong [DY,DX{]}_{\mathrm{st}}$ for finite spectra. Duality reverses a cofibre triangle into a fibre triangle. In particular,This turns a cohomology calculation on a finite object into a homology calculation on its dual, and vice versa. It is Spanier-Whitehead duality, not a claim that the full category of ordinary spaces is self-dual.

Half-exactness is the local image–kernel property; fibres and cofibres, representability, and finite stable duality explain how that property organizes homotopy and cohomology. The distinctions between variance, pointed sets and groups, and finite stable objects and arbitrary spaces are essential to using the duality correctly.

Use finitely generated right projective modules throughout. The group K0 of a ring is the Grothendieck group of their direct-sum monoid, with $[P\oplus Q]=[P]+[Q]$. The group K1 of a ring is ${\mathrm{GL}}_{\infty }(A)/E_{\infty }(A)$, equivalently the abelianization of the stable general linear group. Stabilization is $g\mapsto \mathrm{diag}(g,1)$; $E_{\infty }$ is generated by elementary transvections. The stable elementary subgroup is the commutator subgroup: elementary transvections are commutators using a third index, and stable block interchange and elementary row identities make the quotient abelian.

Write $A_0$ for the lower-right ring, printed as $A^{\prime }$ in the source. A Milnor square has unital commuting maps, is a pullbackand has at least one of $j_1,j_2$ surjective. Interchange the two sides if necessary so that $j_1$ is surjective.

We first establish the patching fact used in the proof. For projective $P_i$ and an isomorphism $\varphi :P_1\otimes A_0\to P_2\otimes A_0$, setFor free modules denote the result by $M_g$. The block matrix identityexpresses $\mathrm{diag}(g,g^{-1})$ as elementary matrices. Their entries lift through $j_1$, and their elementary lifts remain invertible. A change of basis on the first side then trivializes this gluing. Thus $M_g\oplus M_{g^{-1}}\cong A^{2r}$, proving that $M_g$ is finitely generated projective. Notice that we lifted elementary matrices, not arbitrary invertible matrices.

For general $P_1,P_2$, choose $P_1\oplus Q_1\cong A_1^r$ and $P_2\oplus Q_2\cong A_2^s$. The reductions of $Q_1\oplus A_1^s$ and $Q_2\oplus A_2^r$ are isomorphic: insert the common reduced $P_i$ and reorder the two complements. Patch them by such an isomorphism. The sum with $M(P_1,P_2,\varphi )$ now patches two free modules of equal size, so is projective by the free case, making each summand projective. The same free change of basis followed by these direct-summand projections shows that extension to $A_i$ recovers $P_i$. Conversely, write an $A$-projective as the image of an idempotent over a finite free module. Since matrices over the pullback are compatible pairs of matrices, applying the compatible idempotents to the free pullback reconstructs this projective from its two reductions. Compatible module isomorphisms patch as well. This proves Milnor patching of projective modules in the form needed below.

The diagonal maps in the sequence are extension of scalars; the two maps to $A_0$ are differences $j_{1\ast }-j_{2\ast }$. Define the connecting map byIt is unchanged by stabilization or elementary multiplication: an elementary factor lifts through $j_1$ and changes a free basis in the gluing. The identity $\mathrm{diag}(g,h)=\mathrm{diag}(gh,I)\mathrm{diag}(h^{-1},h)$ and the preceding elementary factorization show $M_g\oplus M_h\cong M_{gh}\oplus A^r$ after matching sizes. Thus $\partial$ is an additive homomorphism on $K_1(A_0)$.

Exactness at $K_1(A_1)\oplus K_1(A_2)$. If the two reduced classes agree, choose stabilized representatives $g_1,g_2$ of the same size. Their reductions differ by an elementary matrix $e$ over $A_0$. Lift $e$ to an elementary matrix over $A_1$ and adjust $g_1$ without changing its $K_1$ class. Its reduction now equals that of $g_2$. The compatible pair and its compatible inverse give an invertible matrix over $A$, so the original pair of classes is in the diagonal image. The reverse containment follows from commutativity of the square.

Exactness at $K_1(A_0)$. A gluing obtained from a change of basis over either upper ring patches a free module, so the difference-map image is killed by $\partial$. Conversely, $\partial [g]=0$ means that $M_g$ becomes free after adding a common projective summand. Add a complement of that summand to obtain an actual isomorphism $M_g\oplus A^s\cong A^{r+s}$. Its two local bases express $g\oplus I_s=j_2(v_2)j_1(v_1{)}^{-1}$. Therefore $[g]=j_{2\ast }[v_2]-j_{1\ast }[v_1]$, which lies in the image of the difference map. This uses stable equality of projective classes, not cancellation of arbitrary projectives.

Exactness at $K_0(A)$. Connecting classes reduce to zero on both sides. Conversely, write a class in the kernel as $[P]-[A^r]$ by adding a complement to its negative projective term. Its reductions have the same classes as $A_i^r$. Equality in $K_0(A_i)$ becomes an isomorphism after a finite common free summand, by taking a complement to the auxiliary projective. Use one sufficiently large free summand on both sides. The stabilized $P$ is then locally free with gluing $g$, so the original class is $[M_g]-[A^{r+s}]=\partial [g]$.

Exactness at $K_0(A_1)\oplus K_0(A_2)$. Write a kernel pair as $([P_1]-[A_1^r],[P_2]-[A_2^r])$, using complements and padding to make the free sizes equal. Their reduced positive classes agree. Adding a common finite free summand makes those reductions isomorphic. Patch the resulting projectives, and subtract the matching free $A$-module. Its restrictions are the desired pair. The reverse containment is immediate because both reductions to $A_0$ agree.

These arguments prove the Mayer-Vietoris sequence in algebraic K-theoryNo zero is inserted at either end; this fragment does not assert injectivity at the first term or surjectivity at the last.

For the group ring of the cyclic group, take the cyclotomic Milnor squareThe maps send the cyclic generator to $1$ and to $\zeta$; the lower map sets $\zeta =1$, and the right map reduces modulo $p$. Since ${\Phi }_p(1)=p$, $\mathbb{Z}[\zeta ]/(\zeta -1)={\mathbb{F}}_p$. To verify the pullback, represent an element by a polynomial of degree at most $p-1$. If both evaluations vanish, it is a constant multiple of the minimal polynomial of an algebraic element ${\Phi }_p$, and evaluation at one forces that constant to be zero. Conversely, represent $b\in \mathbb{Z}[\zeta ]$ by $f$ of degree below $p-1$. For a compatible integer $a$, $a-f(1)$ is divisible by $p$; replacing $f$ by $f+[(a-f(1))/p]{\Phi }_p$ yields the pair $(a,b)$. The right map is surjective, so this is a Milnor square.

The hinted cyclotomic units are $u_k=1+\zeta +\cdots +{\zeta }^{k-1}$. If $kl\equiv 1(\mathrm{mod}p)$ with $1\le l<p$, the geometric sum ${\sum }_{j=0}^{l-1}{\zeta }^{kj}$ is their inverse, because ${\zeta }^{kl}=\zeta$. Their residues are $k\in {\mathbb{F}}_p^{\times }$, covering every unit of the common field. Gaussian elimination identifies $K_1({\mathbb{F}}_p)$ with ${\mathbb{F}}_p^{\times }$ by determinant. Thus $K_1(\mathbb{Z}[\zeta ])\to K_1({\mathbb{F}}_p)$ is surjective, and the connecting map vanishes.

The remaining exact sequence injects $K_0(\mathbb{Z}[C_p])$ into $\mathbb{Z}\oplus K_0(\mathbb{Z}[\zeta ])$, with image $\{(r,x):r=\mathrm{rank}x\}$. Here $K_0(\mathbb{Z})=K_0({\mathbb{F}}_p)=\mathbb{Z}$, and reduction of a projective over the integral domain $\mathbb{Z}[\zeta ]$ has dimension its rank: localize at $(\zeta -1)$, where the projective is free, and compare its residue-field dimension with its fraction-field dimension. Projection onto the second factor is therefore a bijection on this kernel. Hence extension along the cyclotomic evaluation givesThe essential step is reduction of cyclotomic units at the ramified prime; surjectivity of a ring map alone would not imply surjectivity on $K_1$.

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.

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.

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.

Write the standard complex coordinates as $z_j=q_j+i{p}_j$ and let $J$ denote multiplication by $i$. Use the standard symplectic form ${\omega }_0={\sum }_{j}d{q}_j\wedge d{p}_j$, for which ${\omega }_0(v,Jv)=|v{|}^2$. In real coordinates $s,t$ on the disk, holomorphicity and the Cauchy-Riemann equations implyTherefore the pulled-back two-form isIf this integral were zero, the continuous nonnegative integrand would vanish everywhere in the interior. The displayed Cauchy-Riemann relation would then make both first derivatives zero, forcing $h$ to be constant on the connected disk and, by continuity, on its boundary. For the nonconstant map under consideration the symplectic area is consequently strictly positive:Take the primitive $\lambda =\frac{1}{2}{\sum }_j(q_{j}d{p}_j-p_{j}d{q}_j)$ from Question 3. Let $c=h{|}_{\partial D^2}$, oriented as the boundary of the disk. Since $c$ lies in $L$, the Stokes theorem givesThe restriction $\lambda {|}_L$ is closed because $L$ is a Lagrangian submanifold. If its Liouville class of a Lagrangian submanifold were zero, it would be an exact differential form, and its integral along every closed curve in $L$ would vanish. The boundary loop above contradicts this. The Liouville class is therefore nonzero. This proves that a holomorphic disk detects a nonzero Liouville class directly by positivity and Stokes; no disk-existence theorem is being assumed beyond the disk given here.