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The dual of the complex tautological line bundle. On the complex projective line the induced frames satisfy $e_1=w{e}_0$ on the overlap of its standard affine charts.

The complex tautological line bundle over Complex projective space isOn the standard chart $U_i=\{[z_0:\cdots :z_n]:z_i\ne 0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The mapis a local trivialization; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial complex line bundle.

Its unit sphere bundle is $S^{2n+1}$: a unit vector $v\in {\mathbb{C}}^{n+1}$ corresponds to $([v],v)$. The projection is the Hopf fibration. Regard $\lambda$ as an oriented real rank-two vector bundle using its complex orientation, and put $t=e(\lambda )\in H^2({\mathbb{CP}}^n;\mathbb{Z})$. The Gysin sequence of a sphere bundle gives, for $2\le q\le 2n$,as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1({\mathbb{CP}}^n)=0$. The CW complex structure has one cell in dimensions $0,2,\dots ,2n$ and none above $2n$, so these groups and products giveThe Euler class of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb{Z}$.

Now let $\alpha$ generate $H^2(S^2;\mathbb{Z})$ and let $u_i$ be its pullback from the $i$th factor of $(S^2{)}^n$. The Künneth theorem gives $H^2((S^2{)}^n;\mathbb{Z})={⨁}_i\mathbb{Z}{u}_i$. If a map $f$ were invariant under all factor permutations, write $f^{\ast }\alpha ={\sum }_i{a}_i{u}_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta :S^2\to (S^2{)}^n$ be the diagonal. ThenThe second condition is $f\circ \Delta =\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. For $n>1$ this is impossible in $\mathbb{Z}$, and $\boxed{\text{no such continuous map exists}.}$ This is the diagonal-degree obstruction to a symmetric sphere retraction.

Use integral cohomology and let $k\ge 1$. On the projective bundle $Y=\mathbb{P}(E)$ define the complex tautological line bundlePut $t=e(\mathcal{L})\in H^2(Y;\mathbb{Z})$, using the canonical complex orientation. On each fibre, $t$ is the Euler class of the tautological line over ${\mathbb{CP}}^{k-1}$, so $1,t,\dots ,t^{k-1}$ restrict to an integral basis of its cohomology.

Here is the finite-cover Leray-Hirsch theorem proof in this case. For each open set $U\subseteq X$, defineIf $E$ is trivial on $U$, its projective bundle is $U\times {\mathbb{CP}}^{k-1}$ and $\mathcal{L}$ is pulled back from the tautological line on the second factor. The Künneth theorem makes ${\Phi }_U$ an isomorphism, since the fibre has finite free integral cohomology. The same holds for every open subset of $U$.

Compactness of $X$ provides a finite trivializing cover $U_1,\dots ,U_N$. Induct on its size. If the result holds on $V=U_1\cup \cdots \cup U_{N-1}$, it holds on $U_N$ and on $V\cap U_N$, both lying in a trivializing chart. Form the diagram of Mayer–Vietoris sequences for the base, with the finitely many degree shifts on the left, and the total space on the right. Naturality of pullback and multiplication by the global even-degree classes $t^i$ makes the diagram commute. The Five lemma gives the isomorphism on $V\cup U_N$. ThusThis is the claimed free module statement, with its graded degree shifts made explicit.

The module basis expresses $t^k$ uniquely using the lower powers, with homogeneous coefficients. Define the Chern classes $c_j(E)\in H^{2j}(X;\mathbb{Z})$ by the unique relationThe pullbacks are suppressed when $f$ is regarded as a polynomial over $H^{\ast }(X)$. Evaluation at $t$ gives a surjective map $H^{\ast }(X)[T]\to H^{\ast }(Y)$. Since $f(T)$ is monic, monic polynomial division over a ring writes any polynomial as $q(T)f(T)+r(T)$ with degree of $r$ below $k$. If its evaluation is zero, module independence forces every coefficient of $r$ to vanish. Hence the kernel is exactly the ideal generated by $f$, provingThe even-degree coefficients are central in the graded commutative algebra, so this division and ideal statement also apply when the base has odd-degree cohomology. Uniqueness of the coefficients proves their naturality under pullback, by pulling back the relation and using the same module basis. With the hyperplane convention $u=-t$, the relation has the usual all-positive Chern coefficients; the alternating signs here correspond to the tautological line itself.

Now suppose $E={⨁}_i{L}_i$. The sections of a projective bundle $s_i:X\to Y$ choose the line $(L_i{)}_x\subset E_x$. They satisfy $s_i^{\ast }\mathcal{L}\cong L_i$, so $s_i^{\ast }t=e_i$ and pulling back the relation gives $f(e_i)=0$.

To obtain the full factorization over the possibly torsion-containing base ring, also use the associated open chartsThey contain the images of the sections $s_i$ and cover $Y$. Projection identifies $\mathcal{L}{|}_{V_i}$ with ${\pi }^{\ast }{L}_i{|}_{V_i}$, so $t-{\pi }^{\ast }{e}_i$ restricts to zero on $V_i$. The long exact sequence of the pair lifts this class to $H^2(Y,V_i)$. The relative cup product of the $k$ lifted classes lies inIts absolute image is ${\prod }_i(t-{\pi }^{\ast }{e}_i)$, so that product vanishes. The polynomial ${\prod }_i(T-e_i)$ is monic of degree $k$ and lies in the kernel of evaluation. Subtracting the monic generator $f(T)$ leaves degree below $k$, and module independence again makes the difference zero. ThereforeThe open-cover argument proves the factorization without a non-zero-divisor assumption on the differences $e_i-e_j$.

The complex tautological line bundle on ${\mathbb{P}}^1$ has fibre at a line $\ell \subseteq {\mathbb{C}}^2$ equal to $\ell$ itself:Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$, $\mathcal{O}(k)=\mathcal{O}(1{)}^{\otimes k}$ for $k>0$, $\mathcal{O}(0)$ to be trivial, and $\mathcal{O}(k)=\mathcal{O}(-1{)}^{\otimes (-k)}$ for $k<0$. These are the tensor powers of the hyperplane line bundle.

On the two given charts, use the holomorphic local frames $e_0(w)=(1,w)$ and $e_1(z)=(z,1)$ of $\mathcal{O}(-1)$. The maps $([1:w],a)\mapsto ([1:w],a(1,w))$ and $([z:1],b)\mapsto ([z:1],b(z,1))$ are its trivializations. On the overlap $z=1/w$,Thus the frame-transition factor from $e_0$ to $e_1$ is $w^{-1}$, while the fibre-coordinate transition from chart 0 to chart 1 is $w$. Specifying both avoids an inverse-convention ambiguity.

For $\mathcal{O}(k)$, let $e_i^{(k)}$ be the induced frames; they satisfy $e_1^{(k)}=w^k{e}_0^{(k)}$. Identifying overlap sections using $e_0^{(k)}$, the Čech cochain groups and Čech coboundary for $k\ge 0$ areHere $\mathcal{O}(\mathbb{C})$ means entire functions in the coordinate of the corresponding chart. Global sections are $\ker \delta$, so $a(w)=w^{k}b(1/w)$, or $b(z)=z^{k}a(1/z)$. Expanding the entire function $a(w)={\sum }_{j\ge 0}{a}_j{w}^j$ shows that $b$ is holomorphic at zero exactly when $a_j=0$ for $j>k$. Thus a global section is determined by a polynomial of degree at most $k$, with basis $1,w,\dots ,w^k$ in the chart-0 frame. Therefore $\boxed{\dim H^0({\mathbb{P}}^1,\mathcal{O}(k))=k+1(k\ge 0)}$.