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Take the Hopf map $\eta :S^3\to {\mathbb{CP}}^1\cong S^2$, given by $\eta (z_0,z_1)=[z_0:z_1]$. The domain has positive-dimensional homology only in degree three, while $H_3(S^2;\mathbb{Z})=0$. In degree two the source group is zero, and all remaining positive-degree source groups are zero too. Hence the induced map on homology is zero in every positive degree.

For nontriviality, use the Hopf fibration $S^1\to S^3\stackrel{\eta }{\to }{S}^2$. It is a fiber bundle: phase multiplication gives each fibre, and choosing a normalized representative on either standard projective chart gives a local trivialization. The theorem invoked is the long exact sequence of homotopy groups of a fibration. Since the universal cover of $S^1$ is contractible, ${\pi }_2(S^1)={\pi }_3(S^1)=0$. Its exact sequence consequently gives an isomorphismThe identity class of $S^3$ is nonzero, since an identity map cannot be null-homotopic while inducing the identity on $H_3(S^3;\mathbb{Z})$. Its image under ${\eta }_{\ast }$ is the class of $\eta$, so that class is nonzero. The Hopf map is not homotopic to a constant despite inducing zero on all positive-degree homology.

False. Take $M=S^4$ and $N=S^3$. Both are closed oriented simply connected manifolds and $4>3$. A cohomological obstruction rules out even a topological fiber bundle.

Suppose a fiber bundle $F\to S^4\to S^3$ existed. The long exact sequence of homotopy groups of a fibration, including its component terms, makes $F$ path-connected. A local trivialization over a small open ball in $S^3$ identifies its preimage, an open subset of $S^4$, with $F\times B^3$. Projection onto $F\times \{0\}$ gives a retraction. An open smooth four-manifold has the homotopy type of a CW complex of dimension at most four, so its rational cohomology vanishes in degrees above four. A retract's cohomology injects into that of the ambient space. Therefore

Use the multiplicative cohomological Serre spectral sequence with rational coefficients. Since $S^3$ is simply connected, there is no nontrivial coefficient action, and only columns zero and three occur on $E_2$. Thus $E_3=E_2$, and the only possibly nonzero differential is $d_3$. Let $u$ be the degree-three base generator. It cannot survive, because the sequence converges to $H^{\ast }(S^4;\mathbb{Q})$ and that ring has zero degree-three part. It has no outgoing differential and its only possible incoming differential is from $E_3^{0,2}$. Hence there is $a\in H^2(F;\mathbb{Q})$ withThe class $a$ is nonzero and, by the degree bound above, has a first vanishing power $a^r=0$ for some $r\ge 2$. Multiplicativity says that $d_3$ is a graded derivation. Since $|a|=2$ is even,But $E_3^{3,2r-2}=H^3(S^3;\mathbb{Q})\otimes H^{2r-2}(F;\mathbb{Q})$, in which $u{a}^{r-1}\ne 0$ by minimality of $r$; also $r$ is nonzero over $\mathbb{Q}$. This is a contradiction. Thus no such fiber bundle exists. This is the special case of a sphere cannot fibre over a lower-dimensional odd sphere required here.

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.

The second assertion is true. Separate the points $p_i$ in the circle direction by a small isotopy if necessary. Cutting $S^1\times S$ at fibers just before and after $p_i\times {\gamma }_i$ exposes two copies of $S$. The surface framing is precisely the stated push-off $p_i^{\prime }\times {\gamma }_i$: the two normal directions give homotopic nonzero normal fields along the curve.

For a single fiber-curve surgery, a cut and regluing of these copies by a Dehn twist changes the curve identified with the transverse meridian by one copy of the surface longitude. In the usual convention that negative surface-framed surgery produces a positive Dehn twist, the filling slope $\mu +\lambda$ implements ${\tau }_{{\gamma }_i}^{-1}$. One can see the coefficient locally by tracing a transverse arc across the twist annulus: it acquires one reverse turn along ${\gamma }_i$, and reversing this cut-and-glue identifies the compressible slope as $\mu +\lambda$. Reversing the circle parameter reverses the monodromy convention, without changing the existence of the fiber bundle.

Every filled block is therefore a product of $S$ with an interval, with a modified endpoint identification. Their cyclic assembly is the mapping torus of the product of these inverse Dehn twists, in the order of the $p_i$ along the oriented circle. It has the original closed topological surface $S$ as fiber, hence fibers over the circle. Intersections between the ${\gamma }_i$ do not obstruct this argument: their twists occur in different fibers, and their ordered product need not commute.

Here is an intrinsic definition that also proves well-definedness. For a complex line bundle $L$, define its First Chern class to be the Euler class of its canonically oriented underlying real rank-two bundle: pull its integral Thom class back along the zero section after forgetting relative support. The complex orientation fixes the sign, so this construction makes no arbitrary choice of generator.

For a rank-$r>0$ complex vector bundle $E$, let $\pi :\mathbb{P}(E)\to X$ be its projective bundle of lines, let $S\subset {\pi }^{\ast }E$ be the tautological bundle, and put $h=-c_1(S)=c_1(S^{\ast })$. On every fiber ${\mathbb{CP}}^{r-1}$, $h$ is the positive degree-two generator. The Leray-Hirsch theorem says that if globally defined cohomology classes restrict to a free basis of the cohomology of every fiber, then multiplication by those classes identifies the cohomology of the total space with a free module over the base. Applied here, it givesThe finite trivializing cover in the question is sufficient for this application: the assertion holds on each trivializing open set by the Künneth theorem, since the fiber has finite free cohomology, and the Mayer–Vietoris sequence and the Five lemma glue it over the finite cover. The same local argument constructs the oriented Thom class used for line bundles. It does not require a choice of classifying map.

There are therefore unique classes $c_i(E)\in H^{2i}(X;\mathbb{Z})$ such thatDefine $c_0(E)=1$ and $c_i(E)=0$ for $i>r$. For rank zero the Total Chern class is $1$. This is the projective bundle definition of Chern classes. Existence and uniqueness follow by expressing $h^r$ in the displayed free module basis, with degrees determining each coefficient. The projective bundle and tautological bundle are intrinsic to $E$, and the line Thom class is uniquely fixed by orientation. Hence the resulting Chern classes do not depend on trivializations or other auxiliary choices. For a line bundle the relation is $h+c_1(E)=0$, agreeing with the original normalization. Pulling back this unique relation also proves naturality. This standard construction and the sum theorem are treated in Vector Bundles and K-Theory, Section 3.1.

The requested result is the Whitney sum formula for Chern classes:Here $c(E)=1+c_1(E)+\cdots +c_r(E)$, and the formula holds for complex vector bundles over a common base. In particular, a trivial bundle has total Chern class $1$.

Now take $B={\mathbb{CP}}^{n-1}$ and let $L\subset B\times {\mathbb{C}}^n$ be its tautological bundle. The standard Hermitian inner product gives the rank-$(n-1)$ complex bundle $Q=L^{\perp }$, with $L\oplus Q\cong {\underline{\mathbb{C}}}^n$. The orthogonal complex line flag manifold in the question is precisely $\mathbb{P}(Q)$: over a first line $\ell$, the second line is any line in ${\ell }^{\perp }$. Local orthonormal frames give this identification as a fiber bundle, with fiber ${\mathbb{CP}}^{n-2}$.

Let $x=c_1(L^{\ast })$ on $B$, also writing $x$ for its pullback to $\mathbb{P}(Q)$. By part 3(a), $H^{\ast }(B;\mathbb{Z})=\mathbb{Z}[x]/(x^n)$. The Whitney sum formula for Chern classes givesLet $S_2$ be the second tautological line on $\mathbb{P}(Q)$, and set $y=c_1(S_2^{\ast })$. Thus $x,y$ are exactly the pullbacks of the positive hyperplane classes from the two projective factors. The projective bundle definition of Chern classes for $Q$ givesTogether with $x^n=0$, this gives a surjective graded ring mapThere are no additional relations. Indeed the second relation is monic of degree $n-1$ in $y$, so polynomial division makes its source free over $\mathbb{Z}[x]/(x^n)$ on $1,y,\dots ,y^{n-2}$. The projective bundle formula for complex vector bundles gives exactly the same free basis on the target. The map sends each basis element to its corresponding basis element, and is therefore an isomorphism. ConsequentlyThis is the cohomology ring of the orthogonal complex line flag manifold. Its additive basis is $x^i{y}^j$ with $0\le i<n$, $0\le j<n-1$. As a symmetry check, multiplying the second relation by $y-x$ gives $y^n-x^n=0$, so $y^n=0$ as expected from the second projection. For $n=2$, every line has a unique orthogonal line, and the relations become $x^2=0$, $x+y=0$, giving the ring of ${\mathbb{CP}}^1$. If $n=1$, the space is empty and the second relation is $1=0$, so the printed formula still gives the zero cohomology ring.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

A Lie group is a group and smooth manifold whose multiplication and inversion are smooth. A Lie algebra is a vector space with a bilinear alternating bracket satisfying the Jacobi identity. For a Matrix Lie group $G\subset GL(m,\mathbb{C})$, a logarithmic chart of a matrix Lie group near $g$ sends $h$ to $\log (g^{-1}h)\in \mathfrak{g}=T_{I}G$; the matrix exponential is its local inverse, and the Baker--Campbell--Hausdorff formula makes the local group operations smooth.

For $B_1,B_2\in \mathfrak{g}$, consider the group commutatorIts logarithm takes values in the vector space $\mathfrak{g}$, and expansion at $(0,0)$ givesThus $\mathfrak{g}$ is closed under the commutator. Bilinearity, alternation, and the Jacobi identity follow from matrix multiplication, so this proves that the Lie algebra of a matrix Lie group has

A principal bundle $\pi :P\to M$ with structure group $G$ is a smooth fiber bundle with a free right $G$-action, each fiber a single orbit, and equivariant local trivializations ${\pi }^{-1}(U)\cong U\times G$.

For the right action of the unitary group on $GL(n,\mathbb{C})$, defineThe matrix $g{g}^{\ast }$ is a positive-definite matrix and a Hermitian matrix, and $q(gu)=q(g)$ for $u\in U(n)$. The polar decomposition of an invertible complex matrix gives the unique factorizationConsequentlyis a diffeomorphism. If $W\subset H(n)$ is a neighborhood of zero and $N=q^{-1}(W)$, then $N$ is an open neighborhood of $I$, it is a union of complete orbits, andMoreover, two matrices have the same value of $q$ exactly when they differ by right multiplication by a unitary matrix. Thus $q$ induces the smooth identificationunder which $\pi$ is projection $H(n)\times U(n)\to H(n)$. This is the general linear group modulo the unitary group, and $\pi$ is in fact a globally trivial principal $U(n)$-bundle.

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.

A principal $G$-bundle is a fiber bundle $\pi :P\to B$ with a free right action of a Lie group $G$ that is locally equivariantly isomorphic to $U\times G\to U$. Each fiber is a single $G$-orbit.

A section of a fiber bundle $\pi :E\to B$ is a continuous map $s:B\to E$ with $\pi \circ s={\mathrm{id}}_B$. It chooses a point of each fibre continuously. A section of a vector bundle is the special case for vector fibres; a projective bundle section chooses a line in each fibre of the underlying vector bundle.

The last-column map $p(A)=A{e}_{s+1}$ has fiber $SO(s)$, included as $\mathrm{diag}(B,1)$. Local completion of a unit vector by the Gram-Schmidt process gives local fiber bundle charts. Consequently it has the homotopy lifting property and the long exact sequence of homotopy groups of a fibration.