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Let P,Q be continuous orthogonal projection matrices of equal rank on a common trivial complex vector bundle, with the displayed bound at every base point. Restricting Q to the image of P is injective: if Qv=0, then $v=(P-Q)v$ contradicts the norm bound unless v=0. Equal dimensions give a continuously invertible vector bundle isomorphism. Compactness applies this observation uniformly to a homotopy of image planes.

A Riemannian metric gives inverse vector bundle isomorphismsIn coordinates they have matrices $g_{ij}$ and $g^{ij}$. Every smooth manifold admits a Riemannian metric by a partition of unity, so its tangent and cotangent bundles are isomorphic as real smooth vector bundles. This isomorphism depends on the metric, not merely on the manifold.

A fiber metric on a smooth vector bundle $E$ gives a smooth orthogonal projection $P:E\to F$ onto any vector subbundle. In a local frame of $E$, let $G$ be the metric matrix and $A$ the columns of a frame of $F$. Then $P=A(A^{t}GA{)}^{-1}{A}^{t}G$, proving smoothness. The complementary subbundle $F^{\perp }=\ker P$ maps isomorphically onto the quotient vector bundle. Explicitly $v\mapsto (Pv,[v])$ is a vector bundle isomorphism $E\to F\oplus (E/F)$, with inverse $(f,[v])\mapsto f+(I-P)v$. This inverse is independent of the representative $v$, but the splitting depends on the chosen metric.

A smooth vector field on a smooth manifold $M$ is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle projection $\pi$, so $\pi \circ X={\mathrm{id}}_M$. In a coordinate chart it has the form $X={\sum }_i{X}^i{\partial }_i$ with smooth component functions $X^i$.

Suppose a smooth vector bundle isomorphism $\Phi :M\times {\mathbb{R}}^n\to TM$ covering the identity is given. For the standard basis $e_1,\dots ,e_n$ of ${\mathbb{R}}^n$, put $X_i(p)=\Phi (p,e_i)$. These are smooth vector fields, and the linear isomorphism on each fibre makes $X_1(p),\dots ,X_n(p)$ a basis of $T_{p}M$.

Conversely, given such a global frame of a vector bundle, defineIt is smooth, covers the identity and is a linear isomorphism on each fibre. Its inverse is smooth as well: in any coordinate chart, the component columns of the $X_i$ form an invertible smooth matrix $C(p)$, and the fibre coordinates of the inverse are $C(p{)}^{-1}v$. matrix inversion is smooth on the invertible matrices. Thus the tangent bundle is trivial exactly when a global frame exists, the defining property of a parallelizable manifold.

For a Lie group $G$, write $L_h(g)=hg$. A left-invariant vector field is a field satisfyingSuch a field is determined by $v=X(e)\in T_{e}G$, since $X(g)=(d{L}_g{)}_{e}v$. Conversely, this formula defines a left-invariant field, by the chain rule and $L_h{L}_g=L_{hg}$.

It also proves smoothness even if smoothness was not initially assumed. Choose a smooth curve $c(t)$ with $c(0)=e$, $c^{\prime }(0)=v$. ThenMultiplication is a smooth map on a Lie group, so differentiating its coordinate functions in the second variable yields a smooth function of $g$. This gives smooth components for $X$ in every local chart.

Choose a basis $v_1,\dots ,v_m$ of $T_{e}G$, where $m=\dim G$, and left translate it. Each $L_g$ is a diffeomorphism, with inverse $L_{g^{-1}}$, so $(d{L}_g{)}_e$ is a linear isomorphism and the translated fields are a basis at every $g$. The explicit parallelization of a Lie group by left translations isIts inverse sends $w\in T_{g}G$ to $(g,(d{L}_{g^{-1}}{)}_{g}w)$. Identifying $T_{e}G$ with ${\mathbb{R}}^m$ gives the required product bundle. The differential facts used are linearity, the chain rule, $d(\mathrm{id})=\mathrm{id}$ and smooth dependence of the differential of a smooth map on its base point.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

The associativity of twisted half-smash products follows from an actual isomorphism of the complement bundles. First work over compact parameter spaces, so that a common choice of flags can be made. Choose source, intermediate and target stages $V_i,W_i,Z_i$ satisfyingFor the two twisted half-smash products and their composite the relevant vector bundles have fibersPull the first two bundles back to $Y\times X$. Since $g_y$ is a linear isometry, the decomposition of $Z_i$ into $g_y(W_i)$ and its orthogonal complement gives the fiberwise vector bundle isomorphismIts inverse is given by the two orthogonal projections. The isomorphism varies continuously with $(y,x)$, and the ranks agree:The one-point compactification of an external orthogonal direct sum gives the smash product of its Thom spaces. Using $g_y$ to identify $g_y({\xi }_x)$ with ${\xi }_x$, the isomorphism therefore givesThe left side is a stage of $(Y\times X){⋉}_{h}E$ and the right side is the corresponding two-step prespectrum presentation of $Y{⋉}_g(X{⋉}_{f}E)$.

These stage comparisons commute with structure maps. Indeed, increasing a stage decomposes a suspension vector into its component in $g_y{f}_x(V_j\ominus V_i)$ and the two residual complement components. In either construction the same source component is pulled back to $V_j\ominus V_i$ and used in the same structure map of $E$; the residual components are left in the same final complement. Iterated orthogonal decomposition gives the same result whether the intermediate complement is separated first or last. They also commute with maps of $E$, since the comparison only rearranges the suspension coordinates.

For clarity about the intervening spectrification, maps out of $LP$ into a genuine indexed spectrum $T$ are precisely compatible prespectrum maps $P\to T$. At stage $i$, adjunction from the Thom coordinate identifies such a map with a continuous familywhere the last identification is the genuine indexed spectrum structure of $T$. Compatibility between stages is exactly compatibility of this family with the source structure maps. Thus maps out of the twisted construction are continuous families of maps over the specified linear isometries. A family of such maps over $g$ out of $X{⋉}_{f}E$ is, by applying this universal property a second time, exactly a family of maps out of $E$ over $g_y{f}_x$. This is the same orthogonal-coordinate correspondence just constructed; it respects the identifications imposed by each spectrification. Consequently it descends from the prespectrum presentations to the associated spectra. Common cofinal refinements remove any incompatibility between the independently chosen flags.

The printed hypothesis does not require $Y$ to be compact. For an arbitrary compactly generated parameter space $Y$, use the same construction on compact test spaces mapping into $Y$. Each such family admits the common flags above, and the comparisons agree under restriction because their formula is the canonical orthogonal projection. They therefore assemble into the continuous global comparison in the compactly generated point-set construction. Equivalently, the preceding universal-property correspondence is valid for arbitrary $Y$ and does not impose a uniform finite target stage on all of it. Passing to cellular models for the derived construction gives the requested natural isomorphism

Yes, the two real smooth vector bundles are isomorphic, although the choice is not canonical. A standard smooth manifold is Hausdorff and second countable, hence admits a smooth locally finite partition of unity. Choose Euclidean Riemannian metrics on coordinate neighborhoods and take their partition-of-unity sum. At each point at least one positive coefficient occurs, so the resulting smooth symmetric tensor $g$ is positive definite.

The resulting musical isomorphism isIt covers the identity of $M$ and is linear on every fibre. In coordinates it is $(x,v)\mapsto (x,G(x)v)$, where $G=(g_{ij})$ is a smooth positive-definite matrix. Its inverse uses the smooth inverse matrix $G^{-1}$, giving the map ${♯}_g$. Thus it is a smooth vector bundle isomorphism.

The Riemannian metric supplies the extra choice: no positive-dimensional real vector space has a linear identification with its dual invariant under all changes of basis. Indeed a scalar change $tI$ acts by $t$ on vectors and $t^{-1}$ on covectors, so an equivariant map $L$ would satisfy $tL(v)=t^{-1}L(v)$ for every $t>0$, forcing $L=0$. An isomorphism therefore exists without being a metric-free canonical one.

A real rank-$d$ vector bundle $p:E\to M$ is a continuous family of $d$-dimensional real vector spaces, locally trivial by homeomorphisms $p^{-1}(U)\cong U\times {\mathbb{R}}^d$ which are linear on every fiber. Transition matrices are continuous maps into ${\mathrm{GL}}_d(\mathbb{R})$. The Grassmannian ${\mathrm{Gr}}_k({\mathbb{R}}^n)$ is the space of $k$-dimensional linear subspaces; its topology can be defined by identifying each plane $W$ with its orthogonal projection matrix $P_W$. Local graph charts identify nearby planes with graphs of maps $W_0\to W_0^{\perp }$.

The tautological bundle is $E_{\mathrm{taut}}=\{(W,v):v\in W\}\to {\mathrm{Gr}}_k({\mathbb{R}}^n)$. In graph charts a basis of $W_0$ gives a continuously varying basis of each graph, proving local triviality. The universal quotient bundle has fiber $Q_W={\mathbb{R}}^n/W$. The map $[v]\mapsto (I-P_W)v$ identifies it continuously with the orthogonal-complement bundle $W^{\perp }$, whose graph charts similarly supply local frames. Thus $Q_{\mathrm{taut}}$ is a rank-$(n-k)$ quotient vector bundle.

For the given $E\to M$, at each $m$ choose a basis of $E_m$ and global sections of a vector bundle realizing its basis vectors, using the evaluation-surjectivity assumption. Those $d$ sections remain independent on a neighbourhood of $m$, since the corresponding determinant in a local trivialization is nonzero at $m$ and continuous. Compactness gives a finite cover by such neighbourhoods. Collect all the selected sections as $s_1,\dots ,s_N$; they span every fiber. They define a fiberwise surjective vector bundle morphismIts kernels have constant dimension $N-d$ and form the kernel bundle of a surjective vector bundle morphism. More concretely, in a local frame of $E$ the map has a continuous full-row-rank matrix $A(m)$ and its kernel projection isThis varies continuously and is unchanged by an invertible change of target frame. The Grassmannian as projection matrices therefore makesa continuous map. The fiber quotient map induced by $F_m$ is an isomorphism ${\mathbb{R}}^N/\ker F_m\to E_m$. The same local matrix formula gives continuously varying right inverses, so these fiber isomorphisms give a vector bundle isomorphism. This proves the classification of vector bundles by a universal quotient bundle:The displayed equality in the classification statement is understood as bundle isomorphism, not literal equality of total spaces.

Finally, the universal quotient bundle need not have a nowhere-zero section. Take ${\mathrm{Gr}}_1({\mathbb{R}}^2)={\mathbb{RP}}^1$. Its quotient identifies with $W^{\perp }$, a Möbius line bundle. Write $W_{\theta }=\mathrm{span}(\cos \theta ,\sin \theta )$, with $\theta$ modulo $\pi$. A section of the complement line has the form $a(\theta )(-\sin \theta ,\cos \theta )$, and being single-valued requires $a(\theta +\pi )=-a(\theta )$. If $a$ is nonzero anywhere, the intermediate value theorem forces a zero between that point and its translate by $\pi$; if it is nowhere nonzero the conclusion is immediate. Hence this rank-one quotient bundle has no nowhere-zero section.

Choose the smooth fiber metric supplied by part (a), and let $P_p:E_p\to F_p$ be orthogonal projection. These projections vary smoothly. Indeed, in a local frame of $E$ write its metric as a positive-definite matrix $G(p)$, and let the columns of $A(p)$ be a frame of $F$. ThenThe matrix $A^{t}GA$ is positive definite, so its inverse is smooth. This formula proves smoothness of the intrinsic projections in every chart. It also gives $P^2=P$ and $\mathrm{im}P=F$.

The orthogonal complement $F^{\perp }=\ker P$ is a smooth vector subbundle: applying $I-P$ to a complement frame gives a frame of its fibers, of constant rank $s-r$, after shrinking the chart. Fiberwise,and the restriction $q{|}_{F^{\perp }}:F^{\perp }\to E/F$ is a smooth fiberwise bijective vector bundle morphism. In local frames its matrix is invertible, with smooth inverse, so it is a vector bundle isomorphism.

More explicitly, the desired orthogonal splitting of a vector subbundle isIts inverse is $(f,[v])\mapsto f+(I-P)v$. Replacing $v$ by $v+w$ with $w\in F_p$ leaves $(I-P)v$ unchanged, so the inverse is well-defined. Both maps are smooth and linear on fibers, and their compositions are identities. The splitting exists globally but depends on the chosen fiber metric.

Let $D_{+}^n$ and $D_{-}^n$ be the two closed hemispheres of the sphere. Use $\gamma$ as the clutching function between the trivial bundles on these hemispheres. The clutching construction definesThe projection to $S^n$ is locally a vector bundle. In a collar of the equator, extend the transition function by keeping it constant in the normal coordinate; this gives explicit vector bundle trivializations across the seam. Since the transition matrices lie in the special orthogonal group, the standard fiber orientations and Euclidean inner products agree on overlaps. Thus this is an oriented rank-$k$ vector bundle with a fiber metric.

For $k\ge 2$, choose a base point $u_0$ on the equator. A constant change of the negative hemisphere frame replaces $\gamma$ by $\gamma (u_0{)}^{-1}\gamma$, so we may suppose $\gamma (u_0)=I$. With $e_k$ the last standard unit vector, define the section obstruction for a clutched vector bundleThe crucial point is that a nowhere-zero section can be normalized using the fiber metric. It is then represented by continuous maps ${\sigma }_{\pm }:D_{\pm }^n\to S^{k-1}$ satisfyingon the equator.

If $f_{\gamma }=0$, the extension-null-homotopy criterion for a sphere extends $u\mapsto \gamma (u)e_k$ to $D_{-}^n$. Take this extension as ${\sigma }_{-}$ and the constant map $e_k$ as ${\sigma }_{+}$. They glue to a nowhere-zero section.

Conversely, suppose such a section exists. Restricting a contraction of $D_{+}^n$ shows that ${\sigma }_{+}{|}_{S^{n-1}}$ is homotopic to a constant. The target sphere is path connected, so the constant can be moved to $e_k$. Multiplying this homotopy pointwise by $\gamma (u)$ gives a homotopy from $\gamma {\sigma }_{+}$ to $\gamma e_k$. The first of these maps equals ${\sigma }_{-}{|}_{S^{n-1}}$ and extends over $D_{-}^n$, hence is null-homotopic. So $\gamma e_k$ is null-homotopic too. A freely null-homotopic based sphere map represents zero in its homotopy group; any change of base point sends the zero class to zero. We have proved both directions:For $n=1$, the same argument reads as the extension criterion for a two-point boundary and uses the trivial set ${\pi }_0(S^{k-1})$ when $k\ge 2$. For $k=1$, the special orthogonal group $SO(1)$ is trivial, so the vector bundle is already a trivial vector bundle; the corresponding constant map to $S^0$ has zero obstruction. These cases require no higher homotopy group calculation.

If $k>n$, then $n-1<k-1$ and the permitted vanishing of lower homotopy groups of a sphere makes $f_{\gamma }=0$. Normalize the resulting nowhere-zero section to a unit section $\sigma$, and setLocal orthonormal frames show that these orthogonal complements form a rank-$(k-1)$ vector bundle. The explicit vector bundle isomorphismhas inverse $w\mapsto (w-⟨w,\sigma ⟩\sigma ,⟨w,\sigma ⟩)$. Orient $V^{\prime }$ so that a positive frame followed by $\sigma$ is positive in $V_{\gamma }$. This is splitting a trivial line from a nowhere-zero section and proves the oriented splittingFinally consider the special orthogonal sphere fibrationThe fiber over $e_{s+1}$ consists exactly of $\mathrm{diag}(B,1)$ with $B\in SO(s)$. Near any unit vector, the Gram-Schmidt process continuously completes it to an oriented orthonormal frame, giving local sections of $p$ and local product fiber bundle charts. In particular $p$ has the homotopy lifting property.

Let $\alpha :S^r\to SO(s+1)$ be based and suppose $r<s$. The composite $p\alpha$ is a based null-homotopic map by the given vanishing of ${\pi }_r(S^s)$. Lift a based null-homotopy of $p\alpha$, beginning at $\alpha$, to a homotopy $G:S^r\times [0,1]\to SO(s+1)$. The lift might initially move its base point $u_0$, but $G(u_0,t)$ always fixes $e_{s+1}$. Right multiplication by $G(u_0,t{)}^{-1}$ corrects this: the homotopystill covers the same sphere homotopy, remains based, and starts at $\alpha$. At its endpoint every matrix fixes $e_{s+1}$, so $\tilde{G}(-,1)$ lies in the included $SO(s)$. Thus every homotopy class in the target is represented by one from $SO(s)$:This proves surjectivity of special orthogonal stabilization below the sphere dimension. For $r=0$, this is the corresponding assertion about path components; the positive-rank special orthogonal groups are connected. The lifting argument also explains the zero term ${\pi }_r(S^s)$ in the long exact sequence of homotopy groups of a fibration.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms $g_x$ on the tangent spaces, equivalently a smooth section of ${\mathrm{Sym}}^2{T}^{\ast }M$ with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse ${♯}_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on $M$, giving a smooth vector bundle isomorphism.

For local flattening of a Riemannian metric, choose a manifold chart neighborhood $V$ around $p$ with closure contained in $U$, and let $h={\varphi }^{\ast }\delta$ on $V$. Choose a smooth bump function $0\le \chi \le 1$, with compact support in $V$ and equal to one on a smaller neighborhood $W$ of $p$. SetThe formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\varphi$ gives an isometry to a Euclidean open subset. Outside $U$ the original metric remains unchanged.

Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.

The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in {\mathbb{R}}^n:|x|>1\}$ with $n\ge 2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The geodesic $\gamma (t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.

Here is the intended compact-core completeness for a Euclidean end. Write $\Phi :M\setminus X\to \{|x|>1\}$ for the end coordinates and assume additionally thatis compact for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a geodesic. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the compact set $K_{R+sT}$. Its velocities also stay in a compact subset of $TM$, since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.

For upward stability of Riemannian completeness, lengths of all curves satisfy $L_{\tilde{g}}\ge L_g$, hence $d_{\tilde{g}}\ge d_g$. A $d_{\tilde{g}}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\tilde{g}$ has a bounded largest matrix eigenvalue, so the coordinate straight segment gives $d_{\tilde{g}}(p,q)\le C|\varphi (q)-\varphi (p)|$. Thus convergence to $p$ is also in $d_{\tilde{g}}$. That distance is complete, and Hopf-Rinow yieldsFor disconnected $M$, apply this argument on each component.

A vector bundle morphism over $M$ is a smooth map $F:E\to E^{\prime }$ with ${\pi }^{\prime }\circ F=\pi$ whose restriction $E_p\to E_p^{\prime }$ is linear for every $p$. A vector bundle isomorphism is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local vector bundle trivializations it is multiplication by an invertible smooth matrix, and its inverse matrix is smooth.

For a diffeomorphism $f:N\to M$, defineThe derivative $d{f}_p$ is a linear isomorphism by the chain rule, since $d(f^{-1}{)}_{f(p)}\circ d{f}_p=\mathrm{id}$. In manifold charts, $F$ is represented by the smooth Jacobian matrix of $f$, so it is a smooth bundle morphism. Its inverse iswhich is also smooth and fiberwise linear. ConsequentlyThis identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.

Under the standard smooth manifold hypotheses, the first answer is yes, although the isomorphism requires a choice. A partition of unity combines Euclidean inner products in manifold charts to produce a Riemannian metric $g$. Nonnegative local weights summing to one preserve positivity, and local finiteness preserves smoothness. The musical isomorphism isIt is smooth and fiberwise invertible, with inverse having local matrix $(g_{ij}{)}^{-1}$. It is therefore a vector bundle isomorphism. There is no preferred such map before a metric or equivalent additional structure is chosen.

The second answer is no. On $S^1=\mathbb{R}/2\pi \mathbb{Z}$, compare the trivial vector bundle $S^1\times \mathbb{R}$ with the Möbius line bundleBoth have rank one. A continuous section of a vector bundle of $L$ is represented on $[0,2\pi ]$ by a continuous function $s$ satisfying $s(2\pi )=-s(0)$. If it were nowhere zero, the intermediate value theorem would prevent its endpoint signs from being opposite. Thus $L$ has no nowhere-zero section. The trivial bundle has the section $1$, and a bundle isomorphism would carry that section to a nowhere-zero section of $L$, a contradiction. Hence

A unit nowhere-zero section $\sigma$ of a real vector bundle with a fiber metric defines a rank-one trivial vector bundle. Its fiberwise orthogonal complement is a subbundle: locally complete $\sigma$ to an orthonormal frame. The vector bundle isomorphism $(v,t)\mapsto v+t\sigma$ has inverse $w\mapsto (w-⟨w,\sigma ⟩\sigma ,⟨w,\sigma ⟩)$. For an oriented bundle, orient the complement so that its positive frame followed by $\sigma$ is positive.

A Lagrangian complement to $TX$ in $TM{|}_X$ is identified with $T^{\ast }X$ by $v\mapsto \omega (\cdot ,v){|}_{TX}$. This produces a vector bundle isomorphism $TX\oplus T^{\ast }X\to TM{|}_X$ that is the identity on $TX$ and carries the canonical pairing $\zeta (u)-\eta (v)$ to $\omega$. A suitably chosen tubular neighborhood map therefore matches the ambient two-forms along $X$, enabling the relative Moser theorem.