Fibrations Fiber Bundles and Homotopy Exact Sequences — Examples

Apply F1 to the based map from one point to $X$. Its mapping-path total space identifies with $PX$ and its fiber over $x_0$ with $\mathrm{\Omega }X$ by F2. The contraction is $D(\alpha ,t)(s)=\alpha ((1-t)s)$, with $D(-,0)=\mathrm{i}\mathrm{d}$ and $D(-,1)=c_{x_0}$; it fixes the constant path and is jointly continuous by the F1 proof. Thus every based cube in $PX$ contracts rel boundary by this formula, so its positive homotopy groups vanish and its component set is a singleton.

For a cube $b(u,t)$ representing a class in $X$, a terminal-point lift is $\tilde{b}(u,t)(s)=b(u,1-(1-t)s)$. It starts at $x_0$ as a path in $s$, ends at $b(u,t)$, is constant when $t=1$ or $u$ is on its boundary, and is continuous by the same interval evaluation/transposition as F1. Its distinguished face is $u\mapsto (s\mapsto b(u,1-s))$. For $n=1$ this is precisely the reverse loop. This directly verifies the orientation rather than assuming a sign-free identification.

For $n\ge 2$, the exact segment ${\pi }_n(PX)\to {\pi }_n(X)\stackrel{\mathrm{\partial }}{\to }{\pi }_{n-1}(\mathrm{\Omega }X)\to {\pi }_{n-1}(PX)$ has zero outer groups by step 1.1. Hence $\mathrm{\partial }$ is both injective and onto. In degree one, exactness shows onto ${\pi }_0(\mathrm{\Omega }X)$, since $PX$ has one component. More explicitly the F3 action is transitive, and its stabilizer is the image of the zero group ${\pi }_1(PX)$; therefore its orbit map, and its composite with loop inversion, are bijections.

A based space is nonempty. If $X$ is one point all groups and component sets in the assertion are trivial; additional components of a general $X$ need not be reached by $PX\to X$ and do not affect its based sequence. The computations explicitly include the constant loop, both time endpoints, and degree one without inventing a group law on a general component set. All lifts and contractions used here are specified formulas, hence choice-free.

Put $a=\mathrm{\mid }{z}_1{\mathrm{\mid }}^2$, $b=\mathrm{\mid }{z}_2{\mathrm{\mid }}^2$. The squared norm of $h$ is $4ab+(a-b{)}^2=(a+b{)}^2=1$, so the formula lands in $S^2$ and is continuous. For $(x,y,z)\in U_N=\{z>-1\}$ define $s_N=(\sqrt{(1+z)\mathrm{/}2},(x-iy)\mathrm{/}\sqrt{2(1+z)})$; its squared norm is $(1+z)\mathrm{/}2+(1-z)\mathrm{/}2=1$ and substitution gives $h(s_N)=(x,y,z)$. On $U_S=\{z<1\}$ use $s_S=((x+iy)\mathrm{/}\sqrt{2(1-z)},\sqrt{(1-z)\mathrm{/}2})$. Multiplication of both complex coordinates by $\lambda \in S^1$ preserves $h$. Over $U_N$ any point of a fiber is uniquely $\lambda s_N$, with $\lambda =z_1\mathrm{/}\mathrm{\mid }{z}_1\mathrm{\mid }$; over $U_S$ use $\lambda =z_2\mathrm{/}\mathrm{\mid }{z}_2\mathrm{\mid }$. These continuous coordinates and their inverse $(b,\lambda )\mapsto \lambda s_N(b)$ or $\lambda s_S(b)$ prove ordinary local triviality and surjectivity. Positive square roots are continuous, for $\mathrm{\mid }\sqrt{r}-\sqrt{s}\mathrm{\mid }\le \sqrt{\mathrm{\mid }r-s\mathrm{\mid }}$ when $r,s\ge 0$.

Put $f_N(z)=\max (0,z+1\mathrm{/}2)$, $f_S(z)=\max (0,1\mathrm{/}2-z)$ and ${\rho }_N=f_N\mathrm{/}(f_N+f_S)$, ${\rho }_S=f_S\mathrm{/}(f_N+f_S)$. The denominator is positive on $[-1,1]$, the two weights sum to one, and their supports are respectively $\{z\ge -1\mathrm{/}2\}\subset U_N$ and $\{z\le 1\mathrm{/}2\}\subset U_S$. This finite partition has precisely the closed-support condition required by F1, including at the two poles and zero-weight boundaries.

We first verify locally the fibre clause used in F7. If $(\cos s,\sin s)=(\cos t,\sin t)$, F8 and F11 give $\sin (s-t)=0$ and $\cos (s-t)=1$. By F9, $s-t=m\pi$ for some $m\in \mathbb{Z}$. F12 gives $\cos ((m+1)\pi )=-\cos (m\pi )$ and $\cos 0=1$, so integer induction in both directions gives $\cos (m\pi )=(-1{)}^m$. Since $\cos (s-t)=1$, $m$ is even and $s-t\in 2\pi \mathbb{Z}$. Conversely F9's $2\pi$-periodicity gives equality whenever the difference lies in $2\pi \mathbb{Z}$. Thus the parametrization has exactly the claimed fibres, independently verifying the affected injectivity input to F7. By F5–F7 the exponential covering $\mathbb{R}\to S^1$ is Hurewicz, with discrete fiber $\mathbb{Z}$. Every based positive-dimensional cube in a discrete space is constant: any two of its points are joined by a straight segment and its continuous image in a discrete set is constant along that segment. Thus all positive homotopy groups of $\mathbb{Z}$ vanish. The contraction $(r,t)\mapsto (1-t)r$ fixes zero and kills every positive homotopy group of $\mathbb{R}$. F2 applied to this covering gives ${\pi }_k(S^1)=0$ for $k\ge 2$. F4 supplies ${\pi }_1(S^1)\cong \mathbb{Z}$.

By steps 1.1–1.2 and F1, $h$ is Hurewicz, and AC is used only through that supplier. F3 gives ${\pi }_1(S^3)={\pi }_2(S^3)=0$. The exact segment $0\to {\pi }_2(S^2)\stackrel{\mathrm{\partial }}{\to }{\pi }_1(S^1)\to 0$ therefore makes $\mathrm{\partial }$ an isomorphism. This conclusion needs no generator sign identification.

For $k\ge 3$, step 1.3 makes both ${\pi }_k(S^1)$ and ${\pi }_{k-1}(S^1)$ zero, so exactness gives that the actual induced map $h_{\ast }$ is an isomorphism ${\pi }_k(S^3)\to {\pi }_k(S^2)$. For $k=3$, F4 gives ${\pi }_3(S^3)=\mathbb{Z}$, hence the claimed value of ${\pi }_3(S^2)$.

None of these spheres or fibers is empty. The endpoint degree $k=3$ uses ${\pi }_2(S^1)=0$, not merely knowledge of its fundamental group; it was established in step 1.3. At the chart poles only the appropriate chart is used, and the partition in step 1.2 excludes the other pole from its closed support. Thus all bundle and homotopy computations are justified, with AC propagated exactly as stated.

The quotient is open: for any open $O\subseteq S^n$, its saturation is $q^{-1}q(O)=O\cup (-O)$, open. For $x\in S^n$ let $U_x=\{y:⟨y,x⟩>1\mathrm{/}2\}$. Then $U_x$ and $-U_x$ are disjoint open sets, and $q^{-1}q(U_x)=U_x\bigsqcup (-U_x)$. Each restriction is bijective onto the open $q(U_x)$ and is open by the same saturation argument for open subsets of $U_x$, hence is a homeomorphism. Such sets cover the quotient. Thus $q$ is a two-sheeted covering and is Hurewicz by F1.

Its fiber is discrete with two points. Every based map of a positive-dimensional cube into this fiber is constant, since each line segment in the cube has connected image and a discrete set has no nonconstant path. All positive fiber homotopy groups vanish. F2 then gives $q_{\ast }:{\pi }_k(S^n)\cong {\pi }_k({\mathbb{R}\mathbb{P}}^n)$ for $k\ge 2$, because both adjacent fiber groups are zero, including ${\pi }_1$ of the fiber at $k=2$.

For $n=1$, regard $S^1$ as complex unit numbers. The map $z\mapsto z^2$ is constant on antipodal pairs and has precisely these fibers: $w^2=z^2$ implies $(w-z)(w+z)=0$. It is onto, since $e^{i\theta }$ has square root $e^{i\theta \mathrm{/}2}$. F4 gives a continuous bijection $v:{\mathbb{R}\mathbb{P}}^1\to S^1$. The source is compact as the quotient image of the compact circle by F5–F6, and the target is Hausdorff, so F5 makes $v$ a homeomorphism. The composite $vq$ sends the generator loop $e^{2\pi it}$ to $e^{4\pi it}$, so F7 gives multiplication by two on $\mathbb{Z}$.

For $n\ge 2$, F3 makes the total space path connected and simply connected. The F2 action of ${\pi }_1({\mathbb{R}\mathbb{P}}^n)$ on the two fiber components is transitive because both are in one total-space component, and its stabilizer is the image of ${\pi }_1(S^n)=0$. Its orbit map is therefore a bijection with a two-element set, so the group has exactly two elements. Its nonidentity element squares to identity: its square cannot equal itself by cancellation, leaving only identity. This identifies the group with $\mathbb{Z}\mathrm{/}2$. A path from $x_0$ to $-x_0$ projects to a loop realizing the nontrivial action; its existence follows already from F3.

The spaces and two-point fiber are nonempty. Step 1.3 is the exceptional endpoint $n=1$; it is not covered by the simple-connectivity input in step 2.1. The higher-group computation in step 1.2 remains valid also for $n=1$. There is no claim here for $n=0$, whose total space is disconnected. Every chart was explicitly specified by its centre and all component assignments are unique; no AC or numerable-bundle theorem was used.

The projection descends by F3 since its two identified boundary values agree. Over $U_1=S^1\setminus \{[0]\}$ use the representative $0<t<1$ and coordinate $u$; inverse representatives are locally continuous by F8, hence continuous by F7. The same local argument applies to the second chart. Over $U_0=S^1\setminus \{[1\mathrm{/}2]\}$ use $-1\mathrm{/}2<\tau <1\mathrm{/}2$ with inverse chart $(\tau ,u)\mapsto [\tau ,u]$ for $\tau \ge 0$ and $[1+\tau ,-u]$ for $\tau \le 0$. The formulas agree at zero by the quotient relation and are continuous by F7. The inverse coordinate on the two prequotient neighbourhoods of the seam is respectively $u$ and $-u$, continuous on their disjoint open pieces; F3 descends it. Restriction of a quotient to the inverse image of an open target is quotient, because open sets there are ambient open and the quotient criterion applies. Thus these are genuine inverse homeomorphisms. On one overlap component the fiber transition is identity and on the other it is $u\mapsto -u$.

We first verify locally the fibre clause behind F9. Equality of $(\cos 2\pi s,\sin 2\pi s)$ and $(\cos 2\pi t,\sin 2\pi t)$ gives $\sin (2\pi (s-t))=0$ and $\cos (2\pi (s-t))=1$ by F10 and F13. F11 then gives $2\pi (s-t)=m\pi$. F14 gives $\cos ((m+1)\pi )=-\cos (m\pi )$ and $\cos 0=1$, so integer induction in both directions gives $\cos (m\pi )=(-1{)}^m$. Since the difference cosine is one, $m$ is even and $s-t\in \mathbb{Z}$; the converse is F11's $2\pi$-periodicity. Thus the values agree exactly on the quotient fibres, independently verifying the affected injectivity input to F9. Now write $c([t])=\cos (2\pi t)$, a well-defined continuous circle coordinate by F9; F12 makes the following arithmetic operations continuous. The functions $f_0=\max (0,c+1\mathrm{/}2)$ and $f_1=\max (0,1\mathrm{/}2-c)$ have positive sum, so ${\rho }_i=f_i\mathrm{/}(f_0+f_1)$ form a finite partition. Their supports lie respectively in $\{c\ge -1\mathrm{/}2\}\subset U_0$ and $\{c\le 1\mathrm{/}2\}\subset U_1$. In particular they avoid the missing chart points even at support boundaries. This supplies the F1 numerating data, and F4 applies under the stated AC.

For $u\in J$, the continuous lift of the circuit is $s\mapsto [s,u]$ in $M$. It begins at $[0,u]$ and ends at $[1,u]=[0,-u]$. The family is jointly continuous in $(u,s)$ by the quotient map, so it gives a transport map, not merely separately selected path lifts. F5 compares this family with any universal lifting function, showing that its endpoint map $R(u)=-u$ represents the F2 transport homotopy class.

The homotopy $K(u,r)=(1-2r)u$ stays in $J$, starts at the identity and ends at $R$, and fixes zero for every $r$. Therefore F15 makes $R_{\ast }$ the identity on every $H_q(J;G)$, for every abelian coefficient group $G$. The based contraction $C(u,r)=(1-r)u$ contracts every based cube in $(J,0)$ rel boundary, so all its positive homotopy groups vanish. The two endpoints $-1,1$ are exchanged by $R$ despite this trivial action on invariants.

The interval and bundle fibers are nonempty; $u=0$ is fixed while $u=\pm 1$ is interchanged. Both circuit endpoints and both homotopy endpoints have been computed. The bundle/chart and reversal calculations are choice-free; the only propagated AC is the invocation of F4 in step 1.2. Thus geometric reversal must not be advertised as nontrivial homology or based homotopy monodromy.

Let $Q:\mathbb{R}\to S^1$ be $Q(t)=e^{2\pi it}$, so $q=Q{\mathrm{\mid }}_{[0,1]}$. We first verify locally the equality-of-values clause in F3 on the full real-line map Q. If $Q(x)=Q(y)$, F4 and F9 give $\sin (2\pi (x-y))=0$ and $\cos (2\pi (x-y))=1$. By F5, $2\pi (x-y)=m\pi$. F10 gives $\cos ((m+1)\pi )=-\cos (m\pi )$ and $\cos 0=1$, so integer induction in both directions gives $\cos (m\pi )=(-1{)}^m$. Since the difference cosine is one, $m$ is even and $x-y\in \mathbb{Z}$. Conversely F5's $2\pi$-periodicity gives $Q(x)=Q(y)$ whenever $x-y\in \mathbb{Z}$. Thus this clause no longer depends on the affected published inference. F3 makes Q continuous and onto; every real coset has a representative in $[0,1]$, so q is continuous and onto. It is also a quotient map: a closed subset $K$ of $[0,1]$ is compact by F6 and F7. Its image is compact since pulling an open cover back by $q$ gives an open cover of $K$, whose finite subcover maps to a finite cover of $q(K)$. The geometric circle is Hausdorff (disjoint sufficiently small Euclidean balls separate distinct points), so F8 makes $q(K)$ closed. Thus $q$ is closed. If $q^{-1}(A)$ is closed, surjectivity gives $A=q(q^{-1}(A))$ closed; with continuity this is the quotient criterion. For the refuted statement only continuity and surjectivity are needed. The path $\beta (s)=(\cos (-\pi s),\sin (-\pi s))=Q(-s\mathrm{/}2)$ is continuous.

Suppose a lift $\mathrm{\ell }:I\to [0,1]$ has $\mathrm{\ell }(0)=0$. For $0<s\le 1$, the equation $Q(\mathrm{\ell }(s))=Q(-s\mathrm{/}2)=\beta (s)$ means $\mathrm{\ell }(s)+s\mathrm{/}2$ is an integer by the locally verified clause in step 1.1. Because $0\le \mathrm{\ell }(s)\le 1$ and $0<s\mathrm{/}2\le 1\mathrm{/}2$, that integer must be $1$, so $\mathrm{\ell }(s)=1-s\mathrm{/}2$. In particular $\mathrm{\ell }(s)\ge 1\mathrm{/}2$ for every $s>0$.

The set $[0,1\mathrm{/}4)$ is a relatively open neighbourhood of $\mathrm{\ell }(0)=0$. By step 2.1 its inverse image is exactly $\{0\}$, which is not open in $I$ since every relative neighbourhood of zero contains positive numbers. This contradicts F2, so no such lift is continuous. F1 therefore excludes both Serre and Hurewicz fibrations. The failure occurs at the initial endpoint, despite the unique possible positive-time lift and the value $\mathrm{\ell }(1)=1\mathrm{/}2$. All spaces are nonempty and all paths were explicit; no AC is involved.

For any ordinary parameter space $Z$, let the initial map be $f(z)=(a(z),b(z))\in E$ and let $h:Z\times I\to [0,1]$ satisfy $h(z,0)=a(z)$. Define $L(z,t)=(h(z,t),\min (b(z),h(z,t)))$. Its coordinates are continuous by F3–F4; the same coordinate check into the Euclidean subspace gives continuity into $E$. Indeed $0\le \min (b(z),h(z,t))\le h(z,t)\le 1$, so its range is in $E$.

The fiber over zero is the singleton $\{(0,0)\}$, while for every $x>0$ the fiber contains the distinct points $(x,0)$ and $(x,x)$ and is homeomorphic to $[0,x]$. Any relatively open neighbourhood of zero in $[0,1]$ contains some $x>0$. A chart over such a neighbourhood would, by F2, identify both fibers with one fixed fiber, forcing a singleton to be in bijection with a set containing two distinct points. This is impossible, so no bundle chart exists at zero.

At time zero, $b(z)\le a(z)$ implies $\min (b(z),a(z))=b(z)$, hence $L(z,0)=f(z)$. Also $pL=h$ at every time. This establishes F1 for every space $Z$, proving $p$ Hurewicz without any universal path-space theorem or choice. It also proves the CGWH test conclusion because the triangle and interval are ordinary compact Hausdorff spaces and their interval cylinders have the same topology.

Empty test spaces in step 1.1 give the empty lift; one-point tests give the same explicit path lift. If $a(z)=0$ then $b(z)=0$ and the lift is $(h(z,t),0)$, so emergence from the collapsed fiber is continuous. If $h=0$, the lift is $(0,0)$; if $h=a$ is constant in time, the formula fixes the original point. At $x=1$ and at both homotopy endpoints the same inequalities remain valid. Thus the singleton/interval change does not obstruct HLP but does obstruct local triviality, as claimed.