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For a smooth map $\varphi :M^{\prime }\to M$ and a connection on a vector bundle $E\to M$, the curvature form of a connection on the pullback vector bundle satisfiesas an endomorphism of $E_{\varphi (p)}$. In a frame of a vector bundle written as a row, with section components written as columns, the connection one-form is $\Omega$ and its curvature form of a connection is $d\Omega +\Omega \wedge \Omega$. The pullback connection has matrix ${\varphi }^{\ast }\Omega$; commutation of the pullback of a differential form with the exterior derivative and wedge product of differential forms proves the identity. No pushforward to an ambient vector field is required: $d\varphi (X)$ is a vector field along a map.

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.

Realize the pullback vector bundle asIts fibre over $p$ is canonically $E_{\varphi (p)}$, with the same vector space operations. If a vector bundle trivialization over $U\subset M$ is ${\tau }_U(v)=(\pi (v),\hat{v})$, give the pullback vector bundle the local trivializationDeclare these maps to be local homeomorphisms and use coordinate charts on $M^{\prime }$ to obtain a smooth atlas on the total space. On overlaps the smooth transition maps are $(p,w)\mapsto (p,h(\varphi (p))w)$, hence smooth and fibrewise linear. They satisfy the same cocycle identities as the original transition maps. This gives a well-defined smooth vector bundle, with smooth projection and rank $r$. It also gives the subspace topology inherited from $M^{\prime }\times E$: inside each product ${\varphi }^{-1}(U)\times {\pi }^{-1}(U)$ the defining condition is the graph of $\varphi$ in the base coordinates. Thus the total space is Hausdorff and second countable, as required for a smooth manifold.

For a smooth section of a vector bundle $s$ over $U$, its pulled-back section is more precisely $p\mapsto (p,s(\varphi (p)))$. If $s$ has component column $u:U\to {\mathbb{R}}^r$ in the original vector bundle trivialization, its component column in $T_U$ is $u\circ \varphi$, which is smooth. The pullback is a smooth rank-$r$ vector bundle, and every local section pulls back smoothly.

The divisor determines the bundle through its local equations. Choose an open cover on which $V$ has a reduced holomorphic defining equation $f_i$, using $f_i=1$ on sets disjoint from $V$. Because $V$ is a complex submanifold of codimension of a submanifold one, its ideal is locally generated by a coordinate; consequently $f_i/f_j$ is a nowhere-zero holomorphic function on each overlap. Inside the sheaf of meromorphic functions on a complex manifold, consider the locally free sheaf of rank oneThese local descriptions agree on overlaps. With $e_i=1/f_i$, its transition functions of a vector bundle are specified byThe ratios satisfy the Čech cocycle condition, and the associated holomorphic line bundle associated to a divisor is $[V]$. The section $1=f_i{e}_i$ is holomorphic and vanishes to order one along $V$. Replacing $f_i$ by $u_i{f}_i$, with $u_i$ a holomorphic unit, replaces the frame by $u_i^{-1}{e}_i$ without changing the subsheaf of meromorphic functions. Refining the cover changes no local sections either. This proves choice independence up to isomorphism rather than merely producing a bundle for one cover.

For a complex manifold of dimension $n$, the canonical bundle isIt is the holomorphic line bundle of holomorphic top-degree differential forms. To compute it on Complex projective space, let $S=\mathcal{O}(-1)$ be the complex tautological line bundle, so that $S^{\ast }=[H]=\mathcal{O}(1)$ is the hyperplane line bundle. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Tensoring the tautological quotient sequence by $S^{\ast }$ gives the Euler sequence on complex projective spaceTaking the top exterior power gives $\det T^{1,0}{\mathbb{P}}^n=\mathcal{O}(n+1)$; taking its dual bundle yieldsHere a negative tensor power means the corresponding positive tensor power of the dual bundle.

The proposed sections over $U$ form a sheaf: compatible maps into $L$ glue uniquely, and their projections remain $\varphi$. Addition and multiplication by holomorphic functions are defined in the fibres of $L$, making this a sheaf of ${\mathcal{O}}_X$-modules. If $e_i$ is a holomorphic local frame of $L$ on $W_i\subset Y$, every section over $U\subset {\varphi }^{-1}(W_i)$ has the unique expressionThus the sheaf is a locally free sheaf of rank one. Its transition functions of a vector bundle are those of $L$ composed with $\varphi$, so the corresponding pullback vector bundle is precisely ${\varphi }^{\ast }L$. This proves invertibility locally and compatibility globally.

For the local blowup of a complex manifold at a point, work on the projective chart $Y_i\ne 0$ and writeAll the incidence equations reduce to $x_j=u{t}_j$. Therefore $(u,(t_j{)}_{j\ne i})$ gives a holomorphic coordinate chart isomorphic to ${\mathbb{C}}^n$. On an overlap with $Y_{\ell }\ne 0$,which is a biholomorphism where $t_{\ell }\ne 0$. The incidence subset is closed in the product of two Hausdorff spaces and is covered by these charts; hence it is a complex manifold of dimension $n$. In each chart the exceptional fibre is $u=0$, so it is a smooth complex submanifold, with its induced projective coordinates givingAway from the origin the inverse of the projection is $x\mapsto (x,[x])$, which is holomorphic; the projection is therefore a biholomorphism outside $E$.

To find the canonical bundle, pull back the nowhere-zero top form on ${\mathbb{C}}^n$. In the chart above, its Jacobian determinant givesThe sign depends only on the ordering of the coordinates. This is a section of $K_X$ with divisor on a complex manifold exactly $(n-1)E$: it has that vanishing order in every exceptional chart and no zeros elsewhere. A nonzero meromorphic section of a holomorphic line bundle with divisor $D$ identifies its holomorphic line bundle with $[D]$, by sending the local generator $1/f_i$ of $\mathcal{O}(D)$ to the nowhere-zero frame obtained by dividing the section by $f_i$. ConsequentlyThis also includes $n=1$, when the local map is an isomorphism and the exponent is zero.

For a general complex manifold, choose a coordinate neighbourhood $U$ of $P$, with $P$ sent to $0$, and replace $U$ by the inverse image of this coordinate neighbourhood in the local blowup of a complex manifold at a point. Glue this space to $M\setminus \{P\}$ along $U\setminus \{P\}$ using the preceding biholomorphism. The resulting charts give a complex manifold $\tilde{M}$ and a holomorphic map $\varphi$; the exceptional fibre is ${\mathbb{P}}^{n-1}$ and the map is a biholomorphism off that fibre. For completeness, the local projection is proper because its inverse image over a compact subset is closed in that subset times compact Complex projective space. This ensures the gluing is Hausdorff: separate points with distinct images downstairs, and points over $P$ inside the local blowup. The atlas is second countable as well.

The top exterior power of the differential defines a global holomorphic bundle map ${\varphi }^{\ast }{K}_M\to K_{\tilde{M}}$. Equivalently, it is a section of $K_{\tilde{M}}\otimes ({\varphi }^{\ast }{K}_M{)}^{-1}$. The same local Jacobian determinant has vanishing order $n-1$ on $E$, and the differential is invertible elsewhere. Applying the holomorphic line bundle associated to a divisor construction gives the canonical bundle formula for a point blowup

The fibers must retain their base labels. In particular, a literal untagged union of fibers in $E$ is not sufficient when $f$ is not injective. Use the pullback vector bundleThis makes precise the indexed-union notation in the PDF. For a vector bundle trivialization ${\Phi }_U(e)=(\pi (e),v)$, defineIt is a fiberwise linear isomorphism, with inverse $(p,v)\mapsto (p,{\Phi }_U^{-1}(f(p),v))$. The transition functions are $g_{VU}\circ f$, hence smooth. They satisfy the same cocycle identities as those of $E$ and give a smooth total space of manifold dimension $\dim N+r$.

One may also see that this is an embedded submanifold of $N\times E$: in a bundle chart the constraint is the graph of the smooth map $f$ in the base coordinates, leaving the $r$ fiber coordinates free. Thus the chart construction agrees with its natural subspace smooth structure. No immersion or surjectivity hypothesis on $f$ is required. Therefore

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

Use the standard complex K-theory Thom class convention. For a rank-$r$ complex vector bundle, its K-theory Euler class is the pullback of the K-theory Thom class along the zero section, namelyHere ${\Lambda }^j$ denotes an exterior power, and a Hermitian metric identifies the dual bundle with the conjugate vector bundle. Choosing ${\lambda }_{-1}(E)$ instead changes the K-theory Euler class by the unit $(-1{)}^r[\det E]$, so its kernel and cokernel computations are unchanged.

For the disk bundle $D(E)$ and sphere bundle $S(E)$, the K-theory Thom class identifies $K^i(D(E),S(E))$ with $K^i(X)$. The exact sequence of this pair becomes the K-theory Gysin sequence of a sphere bundle:The groups are two-periodic by Bott periodicity.

View Real projective space as the circle bundle of ${\gamma }^{\otimes 2}$ over Complex projective space, where $\gamma \to {\mathbb{CP}}^n$ is the tautological bundle. Indeed, $S^{2n+1}/\{\pm 1\}={\mathbb{RP}}^{2n+1}$ maps to $S({\gamma }^{\otimes 2})$ by $[v]\mapsto v\otimes v$; in each fiber squaring identifies exactly $v$ and $-v$.

Use Complex K-theory of complex projective space in the formThe K-theory Euler class of ${\gamma }^{\otimes 2}$ is $1-[\gamma {]}^{-2}$. Multiplying it by the unit $-[\gamma {]}^2$ gives $1-[\gamma {]}^2=2t-t^2$. The K-theory Gysin sequence of a sphere bundle therefore givesUnder the projection $q:{\mathbb{RP}}^{2n+1}\to {\mathbb{CP}}^n$, the pullback vector bundle $q^{\ast }\gamma$ is the complexification of a real vector bundle $L$ of the tautological bundle of Real projective space: the fiber map sends $[v,z]$ to $zv$. Thus $t$ pulls back to $1-[L]$. Put $\mu =[L]-1$. The relation $L^{\otimes 2}\cong 1$ gives ${\mu }^2=-2\mu$, and $t^{n+1}=0$ gives $2^n\mu =0$. HenceFor completeness, these relations give precisely the abelian group $\mathbb{Z}\oplus \mathbb{Z}/2^n$. In the basis $t,\dots ,t^n$, the image of multiplication by $2t-t^2$ is generated by $2t-t^2,2t^2-t^3,\dots ,2t^n$; its relation matrix is triangular with determinant $2^n$, and the relations express every $t^j$ as $2^{j-1}t$. Thus there is exactly one cyclic torsion subgroup of order $2^n$.

If $n\ge 1$ and $a=a_0+a_{1}t+\cdots +a_n{t}^n$ is in the kernel, successive coefficients of $(2t-t^2)a$ give $2a_0=0$, $2a_1-a_0=0$, and so on, up to $2a_{n-1}-a_{n-2}=0$. Hence $a_0=\cdots =a_{n-1}=0$, with $a_n$ arbitrary. For $n=0$ the multiplier is zero on $R=\mathbb{Z}$. In all cases,

Finally let $E=L^{\oplus k}$ with $k\ge 1$, over $B={\mathbb{RP}}^{2n+1}$. Since $L^{\ast }\cong L$, multiplicativity of the K-theory Euler class givesOn $K^0(B)=\mathbb{Z}\oplus (\mathbb{Z}/2^n)\mu$, the image of multiplication by this K-theory Euler class is the subgroup generated by $2^{k-1}\mu$: multiplication by an integer produces every such multiple, while multiplication by $\mu$ only adds multiples of $2^k\mu$. ThereforeMultiplication by $\mu$ is zero on $K^{-1}(B)\cong \mathbb{Z}$, since $2^n\mu =0$ and this abelian group has no nonzero torsion elements. Thus the K-theory Gysin sequence of a sphere bundle gives a short exact sequenceIt splits as an exact sequence of abelian groups because its quotient is a free abelian group. ConsequentlyThe convention $\mathbb{Z}/1=0$ covers $n=0$ and $k=1$. For example, when $k=1$ the sphere bundle is $(S^{2n+1}\times S^1)/\{(v,z)\sim (-v,-z)\}$, a mapping torus of the antipodal map; its Topological K-theory has no nonzero torsion elements, agreeing with the formula. If $k=0$ is allowed, the rank-zero vector bundle has an empty unit sphere bundle, so $K^0(S(E))=0$ instead.

For $x\in S^2\subset {\mathbb{R}}^3$, let $F(x)=x^T$, viewed as a nonzero $1$ by $3$ matrix. This defines a smooth map $F:S^2\to X_{1,3}$. Its pulled-back fiber isso the evident fiberwise identity gives an isomorphism of vector bundles $F^{\ast }{E}_{1,3}\cong T{S}^2$.

If $E_{1,3}$ were a trivial vector bundle, its pullback would be trivial. A trivial rank-two bundle has a nowhere-zero section, whereas the assumed form of the Hairy ball theorem says that the tangent bundle $T{S}^2$ does not. Hence $E_{1,3}$ is nontrivial.

Take bundle charts ${\Phi }_i:E{|}_{U_i}\to U_i\times {\mathbb{R}}^r$ with transition functions $g_{ij}:U_i\cap U_j\to GL(r,\mathbb{R})$. Over $f^{-1}(U_i)$ defineTheir transition functions are $g_{ij}\circ f$, so they give the fibre producta smooth manifold and make its projection to $M$ a rank-$r$ pullback vector bundle.

A local section $s:U\to E$ determines the sectionof $f^{\ast }E$ over $f^{-1}(U)$. It is unique with second component $s\circ f$.

For a smooth map between manifolds $f:M\to N$, the pullback vector bundle $f^{\ast }TN$ has fibre $T_{f(p)}N$ at $p\in M$. A smooth section is a vector field along a map. It allows one to differentiate maps and curves without requiring them to be invertible.