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The assertion is false. Take $m=3$ and $n=2$. For the constant attaching map, the cell attachment gives $S^2\vee S^4$. Its degree-two generator has square zero: restricting the square to either sphere gives zero, and restrictions identify its degree-four cohomology with that of the $S^4$ summand.

For the Hopf fibration $\eta :S^3\to S^2={\mathbb{CP}}^1$, the attachment instead gives ${\mathbb{CP}}^2$. One can verify the attaching map explicitly with the characteristic mapIts interior maps homeomorphically to the complement of ${\mathbb{CP}}^1$; on the boundary it sends $(z_1,z_2)$ to $[z_1:z_2:0]$, exactly the Hopf fibration. By part (a), the degree-two generator of $H^{\ast }({\mathbb{CP}}^2;\mathbb{Z})$ has nonzero square generating degree four. HenceThe additive groups agree, but multiplication distinguishes the attachments.

The dimension condition explains why this example is the relevant one. In general the only positive-degree additive generators lie in degrees $n$ and $m+1$. A potentially nonzero product can only be the square of the degree-$n$ generator, and only when $m+1=2n$. Its coefficient is the Hopf invariant of the attaching map. Here the constant map has invariant zero, whereas the complex Hopf fibration has invariant one.

Choose the Hopf map $\eta$ and orientations so its Hopf invariant is one. The cellular cohomology of $Y_d$ is $\mathbb{Z}$ in degrees $0,2,4$ and zero otherwise. Let $x,z$ be generators in degrees two and four. Precomposing $\eta$ with a degree-$d$ self-map of $S^3$ represents $d\eta$. The induced map of mapping cones is the identity on the two-cell and has degree $d$ on the four-cell. Since the Hopf mapping cone has cup square equal to its top generator, naturality of the cup product gives $x^2=dz$. Equivalently, precomposition scales the Hopf invariant by degree. HenceThe coefficient is $d$, not $d^2$: multiplication in ${\pi }_3(S^2)$ is being used, rather than postcomposition by a degree-$d$ map of the target sphere.

For $X=X_{d,f}$, the only nonzero cellular boundary is $C_4\to C_3$, multiplication by $f$. Thus its cellular cohomology givesLet $x$ again generate $H^2$, and let $z$ be the class of the four-cell cochain. Collapsing the three-sphere gives a map $X\to Y_d$ which pulls the degree-two generator back to $x$ and the degree-four generator back to $z$. Hence $x^2=dz$, now interpreted modulo $f$. Every other product of positive-degree classes vanishes by dimension. This is the cohomology ring of a Hopf attachment with a sphere summand. More explicitly, if $f\ne 0$,If $f=0$, there is additionally a generator $y$ of degree three, with $xy=y^2=yz=0$. The displayed description of the groups uses $\mathbb{Z}/0\mathbb{Z}=\mathbb{Z}$, so the zero case is included. In particular, a nonzero $f$ kills the degree-three cohomology but can leave torsion in degree four.

There are no one-cells, and attaching cells of dimension at least three does not change ${\pi }_1$. Thus $X$ is simply connected. The Hurewicz theorem in degree two gives

To compute ${\pi }_3$, choose a map $g:X\to {\mathbb{CP}}^{\infty }=K(\mathbb{Z},2)$ representing $x$. This Eilenberg–MacLane space can be realized by the classifying space of $S^1$. On $S^2$ the map is the standard inclusion; on $S^3$ it is constant, and it extends over the four-cell since ${\pi }_3(K(\mathbb{Z},2))=0$. Its homotopy fibre $T$ is the pullback of the universal circle bundle $S^{\infty }\to {\mathbb{CP}}^{\infty }$. The map on ${\pi }_2$ is an isomorphism, so the long exact sequence of homotopy groups of a fibration gives ${\pi }_1(T)={\pi }_2(T)=0$ and ${\pi }_3(T)\cong {\pi }_3(X)$. Therefore the Hurewicz theorem identifies the latter with $H_3(T;\mathbb{Z})$.

Over $W=S^2\vee S^3$, the circle-bundle total space isThe first summand is the Hopf fibration total space over $S^2$, and the second is the trivial bundle over $S^3$. Their intersection is the fibre over the wedge point. The Mayer–Vietoris sequence gives $H_3(P)=\mathbb{Z}A\oplus \mathbb{Z}B$, with generators from the two three-spheres. These are the lifts of ${\eta }_1$ and $i_2$. The circle bundle over $W$ also has a 2-connected total space, so lifting and the Hurewicz theorem identify $d{\eta }_1+f{i}_2$ with $dA+fB$.

Over the attached four-disc the bundle is trivial. The resulting relative pair has the homology of $(S^1\times D^4,S^1\times S^3)$, so $H_4(T,P)=\mathbb{Z}$ and $H_3(T,P)=0$. Its boundary map sends a generator to the lift of the attaching map, namely $dA+fB$. The long exact sequence in relative homology now givesApplying Smith normal form to this one relation proves the third homotopy group of a Hopf attachment with a sphere summand:Since $d>0$, the greatest common divisor is positive even when $f=0$. This derivation uses the required map to an Eilenberg–MacLane space and determines the extension, rather than merely the orders of its pieces.

The level equation is $|z_1{|}^2+|z_2{|}^2=|z_0{|}^2$. In particular $z_0\ne 0$, so the global affine coordinates on this level are $w_j=z_j/z_0$. They identify it withThe diagonal circle action becomes $(w_1,w_2)\mapsto (t{w}_1,t{w}_2)$ and is free: at least one coordinate is nonzero, so fixing a point forces $t=1$. Since the circle is compact, the Free proper Lie-group action theorem makes the quotient a smooth two-dimensional smooth manifold.

The map $(w_1,w_2)\mapsto [w_1,w_2]$ is the Hopf fibration. It is onto the complex projective line, and two unit representatives determine the same complex line precisely when they differ by a scalar of modulus one, hence by a circle orbit. Local sections, for example $[1,u]\mapsto (1,u)/\sqrt{1+|u{|}^2}$, show that the induced identification is a diffeomorphism. ThereforeThis identifies the symplectic reduction as a smooth manifold; no computation of its reduced symplectic form is needed.

Two explicit fibrations of $L(2,1)={\mathbb{RP}}^3$ are supplied by Seifert fibrations of real projective 3-space:

Thus these are Seifert fibered descriptions of the same filled manifold with different numbers of exceptional fibers.

A map $p:E\to B$ is a Serre fibration when it has the homotopy lifting property for every disc: given $g:D^k\to E$ and $H:D^k\times I\to B$ with $pg=H(-,0)$, there is a lift $\tilde{H}:D^k\times I\to E$ satisfying $\tilde{H}(-,0)=g$ and $p\tilde{H}=H$.

Fix $b_0\in B$, $e_0\in F=p^{-1}(b_0)$, and write $i:F\hookrightarrow E$. The long exact sequence of homotopy groups of a fibration isThe first two maps are induced by inclusion and projection. To define $\partial$, represent a class of ${\pi }_k(B)$ by a map of pairs $(D^k,S^{k-1})\to (B,b_0)$, lift it beginning at $e_0$ along radial paths, and restrict the lift to $S^{k-1}$; that restriction lies in $F$. At the bottom, exactness continues through the pointed sets ${\pi }_1(B)\to {\pi }_0(F)\to {\pi }_0(E)\to {\pi }_0(B)$.

Every fiber bundle is a Serre fibration. Pull a bundle back along $H:D^k\times I\to B$; a lift of $H$ is the same as a section of this pullback extending the section over $D^k\times \{0\}$ supplied by $g$. Since $D^k\times I$ is compact, finitely many bundle charts cover it. A Lebesgue-number subdivision of $I$, followed by a finite subdivision of $D^k$, makes each resulting prism lie in one chart. In a trivialization $U\times F\to U$, extend the section across a prism by keeping its $F$-coordinate constant along the interval direction. Proceed prism by prism and time-slab by time-slab; on an already treated face use its prescribed coordinate, and the transition functions ensure agreement on overlaps. The resulting sections glue to the required lift $\tilde{H}$.

Now consider $i:{\mathbb{RP}}^n\hookrightarrow {\mathbb{CP}}^n$. The target is simply connected, so $i_{\ast }$ is zero on ${\pi }_1$. For $k\ge 2$, every based map $S^k\to {\mathbb{RP}}^n$ lifts through the double cover to $S^n$. The compositelifts through the Hopf fibration $S^{2n+1}\to {\mathbb{CP}}^n$ to the real-coordinate inclusion $S^n\hookrightarrow S^{2n+1}$. This inclusion is null-homotopic because ${\pi }_n(S^{2n+1})=0$. Hence $i_{\ast }$ is zero on every ${\pi }_k$, including the cases $n=1$ and $k\ge 2$ where the source groups already vanish.

An orientation of a vector bundle of real rank $d$ over $R$ is a coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. For a complex bundle of rank $m$, a complex basis gives the real basis $(v_1,i{v}_1,\dots ,v_m,i{v}_m)$. A complex change of basis has positive real determinant $|{\det }_{\mathbb{C}}A{|}^2$, so these local choices agree and give the canonical orientation of a complex vector bundle over $\mathbb{Z}$, hence over every commutative ring $R$.

Let $D(E)$ and $S(E)$ be the disk and sphere bundles of an $R$-oriented rank-$d$ bundle over compact $B$. A Thom class $u_E\in H^d(D(E),S(E);R)$ restricts to the chosen generator on every fibre. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$ for the zero section $s$. Substituting the Thom isomorphism into the long exact sequence of the pair $(D(E),S(E))$ gives the Gysin sequence of a sphere bundle

Apply this to the Hopf fibration and induct on $n$. If $x=c_1(\mathcal{O}(1))$ has degree two, the result is the cohomology ring of complex projective space

The Künneth theorem over a principal ideal domain gives a natural short exact sequence with tensor-product and Tor terms; the sequence splits, though not naturally. Here all groups are free, so the Tor term vanishes andIf a graded-ring automorphism sends $x$ to $ax+by$, then $(ax+by{)}^{n+1}=0$. The coefficients of the nonzero mixed monomials force $ab=0$; the same applies to the image of $y$. Invertibility then forces a signed permutation matrix on the basis $x,y$. Conversely, complex conjugation on either factor changes the sign of its degree-two generator, and swapping the factors exchanges $x$ and $y$. Thus the realizable group isthe group of all signed $2$ by $2$ permutation matrices, as described by the cohomology automorphisms of a product of two complex projective spaces.

Fibrewise evaluation identifiesThe identity endomorphism is a nowhere-zero continuous section, giving the canonical trivialization of a line bundle tensored with its dual.

Give a complex line bundle its natural orientation as a real plane bundle. Then its Euler class equals its First Chern class. If $a,b$ are the standard generators of $H^2(S^2\times S^2;\mathbb{Z})$, choose complex line bundles $L_1,L_2$ pulled back from the Hopf fibration on the two factors, with $e(L_1)=a$ and $e(L_2)=b$. The triviality of $L_i^{\ast }\otimes L_i$ gives $e(L_i^{\ast })=-e(L_i)$. Hence, for anythe tensor product $L_1^{\otimes p}\otimes L_2^{\otimes q}$, with negative powers interpreted using dual bundles, has Euler class $\alpha$. Its underlying oriented real rank-two bundle is the required $E_{\alpha }$.

Normalize on the unit sphere:This is an $S^1$-equivariant map of the Hopf fibration covering $\bar{f}$, and it has degree one on every circle fiber. The bundle isomorphism in part c givesSince this first Chern class generates the cohomology ring of ${\mathbb{CP}}^{n-1}$, $\bar{f}$ acts identically on its cohomology. Naturality of the oriented Gysin sequence of a sphere bundle then gives $\mathrm{deg}F=1$.

The original homogeneous map is properly homotopic to the cone on $F$ by radial normalization; the norms $\Vert f(v)\Vert$ are bounded above and away from zero on the unit sphere, so this homotopy is proper. Thus $f$ has proper degree one. It fixes the generator of $H_c^{2n}({\mathbb{C}}^n)$, and all other compactly supported groups vanish. Therefore $f^{\ast }:H_c^{\ast }({\mathbb{C}}^n)\to H_c^{\ast }({\mathbb{C}}^n)$ is the identity.

For $\zeta =z_2/z_1$ and $\eta =1/\zeta$, the tautological frames $(1,\zeta )$ and $(\eta ,1)$ differ by $1/\zeta$. Normalize their ambient Hermitian form norms; the resulting frames differ by the displayed phase. The unit circle total space is $S^3\subset {\mathbb{C}}^2$ and its projection to the complex projective line is the Hopf fibration. The transition of the dual line is the reciprocal phase.