advanced26 min studyLesson 479 in Mathematics

Long exact homotopy sequences

Read the fibration sequence degree by degree and handle its low-degree component terms as pointed sets.

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What you’ll understand

  • State the based fibration exact sequence with its connected-base and nonempty-fiber choices.
  • Compute a connecting homomorphism in the Hopf fibration.
  • Interpret low-degree exactness through components and monodromy rather than invented group kernels.

A fibration links three spaces whose homotopy groups constrain one another. The long exact sequence organizes those constraints so that a difficult group can sometimes be recovered from its neighbors. Exactness has a precise meaning: classes that become trivial under one map are exactly those produced by the previous map. Near degree zero, however, the terms need not be groups, and the sequence must be read using distinguished components and actions rather than ordinary abelian algebra.

Choose a fiber and compatible basepoints

Let p:E→Bp:E\to B be a Serre fibration, or a Hurewicz fibration, choose e0∈Ee_0\in E, and put b0=p(e0)b_0=p(e_0) and F=p−1(b0)F=p^{-1}(b_0). Assume B path connected for the displayed tail; otherwise restrict to the path component of b0b_0 and its inverse image. The sequence has maps induced by inclusion and projection, together with connecting maps ∂:πn(B,b0)→πn−1(F,e0)\partial:\pi_n(B,b_0)\to\pi_{n-1}(F,e_0). For n at least two these are group homomorphisms. The based groups of F and E concern the components of their chosen points even if the entire spaces are disconnected.

To picture the connecting map, represent a based sphere by a cube with boundary sent to b0b_0. Lift from prescribed faces using homotopy lifting. The remaining boundary face lands in F and represents the lower-dimensional class. Choices change the representative by a based homotopy, so the resulting class is well-defined. For n equal to one, the lifted loop ends somewhere in F, and the connecting map records its path component. This endpoint-component description is a map to a pointed set, not in general a homomorphism into a group.

⋯→π2(E)→π2(B)→∂π1(F)→π1(E)→π1(B)→∂π0(F)→π0(E)→∗\cdots\to\pi_2(E)\to\pi_2(B)\xrightarrow{\partial}\pi_1(F)\to\pi_1(E)\to\pi_1(B)\xrightarrow{\partial}\pi_0(F)\to\pi_0(E)\to *
The base is path connected and a nonempty chosen fiber supplies basepoints. The π₀ terms are pointed component sets; omitted group basepoints are the compatible chosen ones.↔ Scroll wide equations horizontally.

Exactness changes meaning at the tail

At a group term, exactness says image equals kernel. For example, a base loop lifts to a loop up to a path inside F precisely when its class lies in the image of π1(E,e0)\pi_1(E,e_0). At a pointed-set term, replace kernel by the preimage of the distinguished element. Thus the components of F mapping into the component of e0e_0 in E are exactly the endpoint components reached by lifted loops from e0e_0. There is no reason for the component set of an arbitrary fiber to possess an addition operation.

Monodromy gives a fuller description: π1(B,b0)\pi_1(B,b_0) acts on the path components of F, and its orbits correspond to path components of E when B is path connected. If E is path connected, that action is transitive, and the stabilizer of the distinguished fiber component is the image of π1(E,e0)\pi_1(E,e_0). This statement generalizes the coset description for covering spaces. It does not require F itself to be path connected. Ignoring the action can lose information about components even when every higher group shown in an attempted calculation vanishes.

im⁡(π1(E,e0)→π1(B,b0))=∂−1([F0]),π0(E)≅π0(F)/π1(B,b0)\operatorname{im}(\pi_1(E,e_0)\to\pi_1(B,b_0))=\partial^{-1}([F_0]),\qquad \pi_0(E)\cong\pi_0(F)/\pi_1(B,b_0)
F₀ is the component of e₀; the second quotient denotes an orbit set, not a quotient group.↔ Scroll wide equations horizontally.

Use known zeros without over-solving the sequence

If both neighbors of a connecting homomorphism vanish in the appropriate positions, exactness can force it to be an isomorphism. More generally, a segment 0→A→B→C→00\to A\to B\to C\to0 records a group extension and does not automatically identify B with a direct product. The maps carry information beyond the names of the groups. The path-space fibration, whose total space is contractible, yields πn(B)≅πn−1(ΩB)\pi_n(B)\cong\pi_{n-1}(\Omega B) for n at least two and identifies components of the loop space with based loop classes at the tail.

The Hopf bundle provides a geometric example with circle fiber, three-sphere total space, and two-sphere base. Its projection sends a unit complex pair (z0,z1)(z_0,z_1) to the complex line it spans, identifying the base with CP1≅S2\mathbb CP^1\cong S^2. Fibers are common complex-phase circles, and standard local charts show it is a bundle, hence a fibration in this setting. The exact sequence explains nontrivial higher sphere maps without claiming the total space is a product. Choosing generators and orientations fixes signs in connecting maps; an unspecified orientation gives an isomorphism only up to sign.

The Hopf boundary map is an isomorphism

Use S1→S3→S2S^1\to S^3\to S^2 and the facts π2(S3)=0\pi_2(S^3)=0, π1(S3)=0\pi_1(S^3)=0, and π1(S1)=Z\pi_1(S^1)=\mathbb Z to determine π2(S2)\pi_2(S^2) and its connecting map.

  1. The relevant segment is 0→π2(S2)→∂Z→00\to\pi_2(S^2)\xrightarrow{\partial}\mathbb Z\to0. Exactness at π2(S2)\pi_2(S^2) gives a trivial kernel for the connecting homomorphism.

  2. Exactness at Z\mathbb Z gives surjectivity, since the next map has zero target. Hence ∂\partial is an isomorphism and π2(S2)≅Z\pi_2(S^2)\cong\mathbb Z.

  3. Choose the sphere generator so that its image is the positive circle generator. With this compatible convention ∂(k)=k\partial(k)=k. Reversing one orientation changes the displayed sign but not the isomorphism conclusion.

The result

The connecting map detects an integer two-sphere class through a loop in the fiber; the group is recovered from exactness rather than a product decomposition.

Explore the model

Interactive lab

Interactive lab · modular

Read a component-valued connecting map

Pause and Reset; set modulus to 4 and additive step to 1. Predict the four endpoint components visited by successive lifted base loops in a fourth-power circle cover. Reset, set modulus to 4 and step to 2, and predict the two-component orbit. Multiples of four return to the distinguished component.

Read a component-valued connecting mapAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe residues illustrate the low-degree boundary map for a discrete four-point fiber. That fiber can be labeled by a cyclic group in this example, but π₀ of a general fiber is only a pointed set.

Check your understanding

How should exactness be interpreted at π₀(F) when the fiber is not assumed to have a group structure?

Try it yourself

For p:S1→S1p:S^1\to S^1, p(z)=z4p(z)=z^4, describe the tail Z→p∗Z→∂π0(F)\mathbb Z\xrightarrow{p_*}\mathbb Z\xrightarrow{\partial}\pi_0(F) with the four-point fiber labeled 0,1,2,3 from the chosen start.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The projection multiplies winding by four, while the connecting map records the lifted endpoint sheet.

Compare with the solution

The first map is k↦4kk\mapsto4k. The second sends winding k to the component labeled k mod 4k\bmod4. Its distinguished-component preimage is 4Z4\mathbb Z, exactly the image of p∗p_*. All four components lie in one monodromy orbit because the total circle is connected. The residue labeling is convenient here and does not turn general component sets into groups.

Take it with you

  • A fibration sequence needs a chosen nonempty fiber and compatible basepoints.
  • Connecting maps and exactness constrain group values through their actual maps.
  • The low-degree tail uses pointed sets and monodromy orbits, not assumed group operations.

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