Fibrations Fiber Bundles and Homotopy Exact Sequences

Taking $X=D^0$ in F1 gives path lifting, including constant paths and any initial point that exists. If $X$ is empty the unique empty lift suffices. This does not assert that a lift of a constant path is constant.

Disk HLP also solves a disk-cylinder problem prescribed on its bottom and sides. Here is the geometric change of domain: $D^n\times \{0\}\cup S^{n-1}\times I$ is a boundary disk parametrized by $x\mapsto (2x,0)$ for $\mathrm{\parallel }x\mathrm{\parallel }\le 1\mathrm{/}2$ and $x\mapsto (x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel },2\mathrm{\parallel }x\mathrm{\parallel }-1)$ for $1\mathrm{/}2\le \mathrm{\parallel }x\mathrm{\parallel }\le 1$. This is a homeomorphism from $D^n$, with boundary the top rim. The complementary top disk has the same boundary parametrization. Parametrize the two hemispheres of a sphere by these two disks; the resulting boundary homeomorphism extends radially from an interior point to a homeomorphism of balls. Applying this construction both to the cylinder with its bottom face and to the cylinder with its bottom-and-sides gives a homeomorphism of these pairs. Thus F1 transfers to the required partial domain. For $n=0$ there are no sides and no change is needed.

For the Hurewicz clause use F2 and put $Y=X\times I$. For $u(x)>0$ define $T_s(x,t)=(h(x,s\min (t\mathrm{/}u(x),1)),t-s\min (t,u(x))),$ and for $u(x)=0$ set $T_s(x,t)=(x,t)$. Away from $u=0$ these are continuous formulas. At $a\in A$, F3 and $h(a,v)=a$ give, for any neighbourhood $O$ of $a$, a neighbourhood $V$ with $h(V\times I)\subseteq O$; the second coordinate changes by at most $u(x)$. Hence continuity holds across $u=0$, even at $t=0$. We have $T_0=\mathrm{i}\mathrm{d}$ and $T_s{\mathrm{\mid }}_W=\mathrm{i}\mathrm{d}$. At $s=1$, either $t\le u(x)$ and the height is zero, or $t>u(x)$, whence $u(x)<1$ and $h(x,1)\in A$. Thus $r=T_1:Y\to W$ is a retraction.

Induct over dimensions of the cells of $X$ outside $A$. On each characteristic disk the already prescribed data are exactly bottom-and-sides data; step 1.2 extends them. They agree on attaching boundaries, so descend to each skeleton. AC in F7 selects the extensions for unrestricted cell families, including the successive dimensions; a finite CW pair requires only finitely many choices. Continuity on all of $X\times I$ follows without assuming an ordinary infinite product/colimit interchange: transpose the constructed function to $X\to C_0(I,E)$. Its restriction to every characteristic disk is continuous by F4; F5 makes the transpose continuous, and F4 uncurries it. The initial and subcomplex values are unchanged at every stage. Taking $A=\mathrm{\varnothing }$ proves the CW-test implication; conversely every disk is a CW complex.

Put $w(x,t)=\min (u(x),t)$, whose zero set is $W$, and define $K:Y\times I\to Y$ by $K(y,v)=T_{1-\min (v\mathrm{/}w(y),1)}(y)$ when $w(y)>0$, and $K(y,v)=y$ on $W$. F3 applied to the fixed tracks $T_s(z)=z$ proves continuity at $w=0$; elsewhere it is composition of continuous maps. In particular $K(y,0)=r(y)$, $K(y,w(y))=y$, and $K(z,v)=z$ for $z\in W$.

Apply ordinary HLP in the chosen category once, with parameter $Y$, initial map $fr$, and base homotopy $gK$. It supplies $L$ with $L(y,0)=f(r(y))$ and $pL(y,v)=gK(y,v)$. Then $\mathrm{\ell }(y)=L(y,w(y))$ is continuous, $p\mathrm{\ell }=g$, and $\mathrm{\ell }(z)=L(z,0)=f(z)$ for $z\in W$. This proves the full prescribed relative lift. When $A=X$, $w=0$ and the construction returns $f$; when $A=\mathrm{\varnothing }$ it still applies and ordinary HLP already suffices. The only selections here are one NDR witness and one HLP witness, not an indexed family, so no AC is used. In particular no regular lifting function is assumed.

Write $v=(v_A,v_E)$. Then $p{v}_E=g{v}_A=gH(-,0)$. By F1 the base homotopy $gH$ has a lift $L:X\times I\to E$ with $L(-,0)=v_E$ and $pL=gH$. This uses exactly the original test class: every space for ordinary Hurewicz, every CGWH space for its CG version, or every disk for Serre.

Set $\tilde{H}(x,t)=(H(x,t),L(x,t))$. Its coordinates are continuous, it lands in the pullback by $pL=gH$, and F2–F3 give continuity. In CGWH this is the same categorical pairing into the kified pullback; the CG source property gives the kified target map. Its initial value is $(v_A,v_E)=v$, and $q\tilde{H}=H$. Thus it solves the required HLP problem.

Empty parameter spaces have the empty lift; if the pullback is empty, every allowable initial-map problem has empty parameter space. Disk dimension zero is included in step 1.1. Constant base maps, point bases, $t=0$ and $t=1$ satisfy the same equations, with no uniqueness or regularity needed. Only one existential HLP witness is used; no indexed selections are made. This proves the claim for both types.

Direct evaluation gives $p_f{j}_f=f$ and $r_f{j}_f={\mathrm{i}\mathrm{d}}_X$. The formula $D((x,\gamma ),t)=(x,s\mapsto \gamma ((1-t)s))$ is continuous by F3 applied to its adjoint. It remains in $E_f$ because its path starts at $f(x)$, begins at $(x,\gamma )$ and ends at $j_f{r}_f(x,\gamma )$. It fixes every constant path. Thus $j_f$ and $r_f$ are homotopy inverses, even with a strong deformation retraction onto $j_f(X)$.

For the given HLP problem define ${\eta }_{z,t}:I\to Y$ by ${\eta }_{z,t}(s)={\gamma }_z((1+t)s)$ if $(1+t)s\le 1$, and ${\eta }_{z,t}(s)=H(z,(1+t)s-1)$ if $(1+t)s\ge 1$. Both domains are closed and cover $Z\times I\times I$. At their intersection the values are ${\gamma }_z(1)=H(z,0)$, so F3–F4 prove the joint continuity of the adjoint. All arguments of $H$ lie in $[0,t]$. In particular no division by $t$ occurs at $t=0$.

Transpose step 1.2 and put $\tilde{H}(z,t)=(x(z),{\eta }_{z,t})$. The path begins at ${\gamma }_z(0)=f(x(z))$, so this lands continuously in $E_f$. At $t=0$ it is exactly $(x(z),{\gamma }_z)$, and $p_f\tilde{H}(z,t)={\eta }_{z,t}(1)=H(z,t)$, including $t=0$ by compatibility. Hence F2 holds for every parameter space, and the same adjoint formulas establish the CG version.

Empty $X$ makes all initial-map test domains empty. A one-point $X$ gives the usual based path space; a one-point $Y$ reduces the deformation to the identity of $X$. Both deformation endpoints and both path endpoints were checked above. Existence of a path in $E_f$ forces its final point into a component meeting $f(X)$, explaining the absence of a surjectivity claim. Every operation was a formula, not a path selection. Together steps 1.1 and 2.1 establish the factorization.

A relative representative $a$ has $pa=b_0$ on its entire boundary because $a(t=0)\subseteq F$ and $a{\mathrm{\mid }}_J=e_0$. Relative homotopies similarly project to based homotopies. Thus composition defines the displayed pointed map; for $n\ge 2$ it preserves coordinate-one concatenation by F2. Given an absolute representative $b$ in $B$, F1 constructs a lift constant on $J$, hence a relative representative with projection exactly $b$. This proves surjectivity in every degree, including $n=1$.

Suppose relative representatives $a_0,a_1$ have based-homotopic projections, through $b(u,t,r)$, $r\in I$. Treat $(u,r)\in I^{n-1}\times I$ as the parameter disk and lift in reversed $t$-time. At $t=1$ prescribe $e_0$ everywhere. On the parameter boundary prescribe $a_0(u,t)$ at $r=0$, $a_1(u,t)$ at $r=1$, and $e_0$ on $u\in \mathrm{\partial }{I}^{n-1}$. These prescriptions agree on intersections, and project to $b$ there because the base homotopy is based. F3 extends them over the full cylinder. At $t=0$ the lift lies in $F$, since $b(u,0,r)=b_0$; on every $J$ face it is $e_0$. It is therefore a relative homotopy from $a_0$ to $a_1$. For $n=1$ the parameter disk is just the $r$ interval and its two endpoints carry the two given paths, so the same argument proves injectivity of pointed sets.

Steps 1.1–1.2 prove bijectivity. The connecting map equals the relative boundary map of F2 composed with the inverse bijection. Consequently both arbitrary lift choices and representative changes give the same output. For $n\ge 2$ a bijective homomorphism has a homomorphic inverse, so this composite is a homomorphism. For $n=1$ it is a pointed map: the constant base representative admits the constant $e_0$ lift, whose initial component is distinguished.

Specifying $e_0$ excludes empty fibers and empty total/base spaces; zero-dimensional boundary cubes when $n=1$ record points, not a nonexistent relative ${\pi }_0$. Constant cubes and point spaces are included by the same construction. All extension domains above are finite cubes with finite subcomplexes; F3 therefore uses only finitely many existential witnesses, and no AC. The equations at $t=0,1$ and $r=0,1$ establish every required endpoint condition.

Replace each relative term ${\pi }_n(E,F)$ of F2 by ${\pi }_n(B)$ using F1. The composite from ${\pi }_n(E)$ is actual composition with $p$, since the relative representative is the same cube. The outgoing map is exactly ${\mathrm{\partial }}_p$ by F1. Bijections preserve inverse images of distinguished points and images; in group degrees F1 gives group isomorphisms. F2 therefore proves the displayed exactness through ${\pi }_0(F)$, including ${\pi }_1(B)$ with a pointed-set outgoing map.

A component of $E$ containing a fiber point maps to $[b_0]$. Conversely, if $p(e)$ is joined to $b_0$, lift a path from $p(e)$ to $b_0$ starting at $e$ using F4. Its endpoint is in $F$ and in the component of $e$. This proves exactness at ${\pi }_0(E)$. The image of the last arrow is by definition the set of components meeting $p(E)$, proving the precise surjectivity criterion.

To verify the proposed action, first fix a path $\gamma$ and two lifts whose initial points are joined by a path in its initial fiber. More generally let the base paths vary through an endpoint-fixed homotopy. On a square prescribe the two given lifts on its vertical sides and the given initial fiber path on its bottom. F4 extends the lift, using base path time as lifting time and the other coordinate as parameter. The top is a path in the terminal fiber, so the endpoint components agree. This proves simultaneous independence of the initial point within its component, of the lift, and of the endpoint-fixed base representative. Existence is path lifting.

The constant lift proves the identity law on components. Pasting a lift of $\gamma$ with a lift of $\eta$ starting at its endpoint gives a lift of $\gamma \ast \eta$. Step 1.3 therefore proves $([e]\cdot [\gamma ])\cdot [\eta ]=[e]\cdot ([\gamma ][\eta ])$. Reversal gives inverses. If a loop in $E$ is based at $e_0$, it exhibits that its projected loop stabilizes $[e_0]$. Conversely, for a stabilizing base loop, lift it from $e_0$ and join its endpoint to $e_0$ in $F$. Pasting gives a based loop of $E$ whose projection is the original loop with a constant segment, hence the same based class. Thus the stabilizer is exactly the claimed image.

Points related by the action are connected in $E$ by a lifted loop, with possible fiber paths. Conversely a path in $E$ between two points of $F$ projects to a loop at $b_0$ and witnesses the corresponding action relation. Thus orbits are exactly fibers of $i_{\ast }$, even when $E$ or $F$ is disconnected. A lift of $\gamma$ ending at $e_0$, read backwards, is a lift of ${\gamma }^{-1}$ starting there. Its initial component is therefore $[e_0]\cdot [\gamma {]}^{-1}$ by step 1.3, proving the boundary formula.

The specified $e_0$ makes $E,B,F$ nonempty, but other base components may have empty fibers; step 1.2 explicitly allows this. In degree one the boundary is a component and no group structure on ${\pi }_0(F)$ is asserted. For a one-point or path-connected fiber its component action is trivial. All extra lifting problems use a point or finite square, and uniqueness of the resulting component defines the action without choosing a family of lifts. The preceding steps prove exactness, action laws, and all low-degree qualifications.

If $e\in F$, then $p^{\mathrm{\prime }}u(e)=vp(e)=v(b_0)=b_0^{\mathrm{\prime }}$, so restriction defines the continuous based $u_F$. The identities $ui=i^{\mathrm{\prime }}{u}_F$ and $p^{\mathrm{\prime }}u=vp$ imply commutation of the inclusion and projection squares in all degrees by F3, including maps of component sets.

For a based cube $b$ in $B$ choose a lift $a$ constant at $e_0$ on its $J$ faces. Then $ua$ lifts $vb$ and is constant at $e_0^{\mathrm{\prime }}$ on those faces. On the distinguished face its restriction is $u_F$ composed with the restriction of $a$. F2 therefore gives ${\mathrm{\partial }}_{p^{\mathrm{\prime }}}{v}_{\ast }[b]=u_{F\ast }{\mathrm{\partial }}_p[b]$. In degree one this is equality of the components of the initial endpoints, so the same calculation covers that square.

If $a$ lifts a loop $\gamma$ beginning at $e\in F$, then $ua$ lifts $v\gamma$ beginning at $u_F(e)$ and ends at $u_F(a(1))$. The independence of component lifts in F1 therefore gives the action identity on every component, without selecting a compatible family of lifting functions.

Steps 1.1–1.3 establish all arrows and the action. Basepoints exclude empty based fibers but no connectedness or surjectivity is needed; components not meeting the image of $u_F$ cause no change in the formulas. For point spaces and constant cubes the equations remain identities; degree zero stays a pointed-set assertion. Every choice above is one representative or lift for one equality, hence no AC.

For the Serre assertion let $q:P={\gamma }^{\ast }E\to I$ and fix $i=0$ or $1$. By F2 it is Serre. Given a based cube $a:(I^n,\mathrm{\partial }{I}^n)\to (P,e)$ with $e\in F_i$, deform its height by $K(x,s)=(1-s)q(a(x))+si$. F3 lifts this finite relative problem starting at $a$, with constant value $e$ prescribed on $\mathrm{\partial }{I}^n\times I$. The endpoint is a based cube in $F_i$ and the lift is a based homotopy in $P$. Thus inclusion is surjective on every ${\pi }_n$, $n\ge 1$. To prove injectivity, start with a based homotopy $a:I^n\times I\to P$ between cubes in $F_i$ and deform its height by the same formula. Prescribe $a$ unchanged on $(\mathrm{\partial }{I}^n\times I)\cup (I^n\times \{0,1\})$, where the height is already $i$. This is a finite cubical subcomplex, so F3 supplies the relative lift; its deformation endpoint is a homotopy wholly in $F_i$. Inclusion is injective, and is a homomorphism because postcomposition preserves cubical concatenation. For components, lift a height path from any point of $P$ to $i$; deform a path between two points of $F_i$ by the same relative procedure with both endpoints fixed. This proves surjectivity and injectivity on ${\pi }_0$. If a fiber is empty then path lifting makes $P$ empty, with the unique bijection of empty component sets and no based assertions. Only finite lifting problems were used.

We first compare continuous lift families. Suppose $a_0,a_1:Z\times I\to E$ project to the two ends of a continuous base-path homotopy $h:Z\times I\times I\to B$, fixed in its path endpoints, and their initial values agree. On the parameter square prescribe $a_0,a_1$ on its two vertical sides and their common initial map on the bottom. The bottom-and-sides square is carried to a single bottom edge by the disk-pair homeomorphism in F3. Tensor this homeomorphism with $Z$ and apply Hurewicz HLP with parameter $Z\times I$ to obtain a lift over the square. Its top edge is a homotopy between the two endpoint maps into the same endpoint fiber (or fiber over the varying endpoint map of $Z$). This argument works for arbitrary ordinary $Z$, and in CGWH with k-products. It needs neither a cofibration of a point in $Z$ nor a regular lifting function.

For a counterexample put $C=\{0,1{\}}^{\mathbb{N}}$ with the topology of finite coordinate cylinders. Every singleton is closed, distinct points are separated by complementary clopen coordinate cylinders, and no singleton is open: any finite cylinder permits changing a later coordinate. Every path in $C$ is constant, since each coordinate of a path maps the connected interval continuously into the discrete two-point space. Likewise every map from $D^n$ into $C$ is constant, by restricting it to line segments between pairs of points (also for $n=0$). Give the set $E=C\times I$ the topology generated by product opens and $V_c=\{c\}\times [0,1)$ for each $c$. The projections $p:E\to I$ and $r:E\to C$ are continuous. Each vertical map $s_c(t)=(c,t)$ is continuous: the inverse image of $V_d$ is $[0,1)$ if $d=c$, otherwise empty, and product opens also have open inverse images.

Fix $n\ge 1$ and a path-connected fiber $F=F_b$ such that ${\beta }_l=\mathrm{i}\mathrm{d}$ on ${\pi }_n(F,e)$ for every loop $l$ at $e$. For any map $f:F\to F$ and path $a:e\to f(e)$ set $A_f={\beta }_a{f}_{\ast }$. This is independent of $a$: if $a^{\mathrm{\prime }}$ is another choice then $a\ast {\bar{a}}^{\mathrm{\prime }}$ is a loop at $e$, so F4 gives ${\beta }_a{\beta }_{{\bar{a}}^{\mathrm{\prime }}}=\mathrm{i}\mathrm{d}$ and hence ${\beta }_a={\beta }_{a^{\mathrm{\prime }}}$. If $H:f\simeq g$ has track $d:f(e)\to g(e)$, F4 gives $f_{\ast }={\beta }_d{g}_{\ast }$; thus correction by $a$ for $f$ equals correction by $a\ast d$ for $g$. Finally the radial-shell formula in F4 commutes pointwise with any continuous postcomposition $g$, giving $g_{\ast }{\beta }_a={\beta }_{ga}{g}_{\ast }$ with the appropriate source basepoints. For corrections $a_f:e\to f(e)$ and $a_g:e\to g(e)$, the concatenation $a_g\ast g(a_f)$ goes from $e$ to $gf(e)$, and this naturality and F4 give $A_{gf}=A_g{A}_f$. All endpoint paths exist by path connectedness, and the resulting maps are uniquely specified, so no indexed choice of paths is required.

Apply step 1.2 with $Z=F_b$, constant initial map in the path parameter equal to $e\mapsto e$, and either two lifting functions for the same path or lifts of two endpoint-fixed-homotopic paths. This proves both independence assertions. The constant path has the continuous constant lift $a(e,t)=e$, so comparison gives $T_{c_b}\simeq \mathrm{i}\mathrm{d}$. Concatenate the chosen continuous lifts of $\gamma$ and $\eta$, the latter starting at the first endpoint. The resulting endpoint is $T_{\eta }{T}_{\gamma }$; comparison with a lift of $\gamma \ast \eta$ proves the composition formula.

In the example, for $f:D^n\to E$ and a base homotopy $H$ starting at $pf$, the map $rf$ is a constant $c$. Consequently $L(z,t)=s_c(H(z,t))$ is a continuous lift with the required initial value. This proves Serre HLP in every degree without choosing any family of lifts. The fiber at $0$ is discrete because the sets $V_c$ isolate its points; the fiber at $1$ has exactly the original cylinder topology of $C$, because every additional generator misses it. Both fibers have only constant paths. Hence any homotopy into either is pointwise constant; homotopy-inverse maps would therefore be inverse homeomorphisms. Such a homeomorphism cannot exist since one fiber is discrete and the other has no isolated points. This refutes unrestricted Serre homotopy equivalence. More precisely, $f:C\to E$, $f(c)=(c,1)$, is continuous, but $H(c,t)=1-t$ has no continuous lift starting at $f$. A lift must have constant label on each time path, hence be $(c,1-t)$; at any fixed $t>0$ the inverse image of $V_d$ is the nonopen singleton $\{d\}$. Thus whole-fiber HLP fails. This counterexample is asserted in ordinary spaces; no compact-generation claim is needed.

The retracing loop $\gamma \ast \bar{\gamma }$ contracts rel endpoints by the formula $\gamma (2t(1-s))$ on $t\le 1\mathrm{/}2$ and $\gamma (2(1-t)(1-s))$ on $t\ge 1\mathrm{/}2$. Reversing the path gives the analogous contraction at $c$. Step 2.1 gives both inverse homotopies displayed in the statement, so transports are homotopy equivalences. If one fiber is empty, a nonempty other fiber would lift the reversed path into it, a contradiction; thus both are empty, and their unique maps are equivalences.

Let $q:P\to I$ be the pullback along $\gamma$, a Hurewicz fibration by F2. For $t\in I$ put ${\alpha }_t(s)=st$ and ${\bar{\alpha }}_t(s)=(1-s)t$. Transport in $q$ gives continuous maps $U:F_0\times I\to P$, $U(e,t)=T_{{\alpha }_t}(e)$, and $V:P\to F_0\times I$, $V(z)=(T_{{\bar{\alpha }}_{q(z)}}(z),q(z))$. These maps are over $I$. Their composites transport along ${\alpha }_t\ast {\bar{\alpha }}_t$ and ${\bar{\alpha }}_t\ast {\alpha }_t$, up to the comparison of step 1.2, now with the extra $t$ or $z$ parameter. Retracing contracts these paths with their endpoints fixed, continuously in the parameter. Step 1.2 therefore gives homotopies of both composites to the identity over $I$, proving the fiber homotopy equivalence. This uses continuous families, not separate equivalences selected for each $t$.

Constant paths, $t=0$ in step 3.2, and one-point fibers satisfy the same comparison argument; it does not require $T_{c_b}$ to equal the identity pointwise. The homology local system and its isomorphisms in F1 now follow by functoriality and homotopy invariance. All constructed homotopies are obtained from single HLP problems, so no indexed choice is used. The argument uses parameter spaces as large as the whole fiber; disk HLP alone would not authorize it. The Serre conclusion follows from its separate finite-domain argument.

Apply step 1.4 to the transport maps. Step 2.1 identifies transports for homotopic base paths and different lifting functions up to homotopy, and gives $T_{\gamma \ast \eta }\simeq T_{\eta }{T}_{\gamma }$. Therefore $A_{\gamma \ast \eta }=A_{\eta }{A}_{\gamma }$ and $A_{c_b}=\mathrm{i}\mathrm{d}$. The reversed loop supplies the inverse. Thus $x\cdot [\gamma ]=A_{\gamma }x$ is a right group action, with automorphisms of ${\pi }_n(F,e)$, independent of all auxiliary endpoint paths and lifts. For $n=1$ the trivial-loop-transport hypothesis says the group is abelian, by the conjugation formula in F4; a simply connected fiber satisfies the hypothesis in every positive degree. The empty fiber has no based group and is excluded by the chosen basepoint; a singleton satisfies the condition and has the trivial action.