Basepoints and induced homomorphisms

Try

A continuous map transforms every loop in its source into a loop in its target. If two source loops deform into one another, composing that deformation with the map gives a target deformation. This makes the fundamental group useful as an invariant: a proposed geometric map must respect the algebra of loops.

Predict

Predict whether homotopic maps always induce literally equal homomorphisms without mentioning basepoints. Their basepoint images may move during the homotopy, so the target groups may even have different basepoints. The correct comparison uses the path traced by the basepoint and a conjugation isomorphism.

Experiment

Follow the paths needed to move a loop based at x₁ to x₀.

Segment	Start	Finish
c	x₀	x₁
α	x₁	x₁
reverse c	x₁	x₀

Concatenate two transported loops. The return path of the first and outward path of the second cancel up to endpoint homotopy. Write the resulting word and verify that it is the transported concatenation. This endpoint check prevents a common reversed-conjugation error.

Understand

A based map f:(X,x₀)→(Y,y₀) is continuous and satisfies f(x₀)=y₀. Define f_*[α]=[f∘α]. This is well defined because composing a homotopy with f preserves its endpoint conditions. It respects multiplication because f applied to a concatenation is the concatenation of the images with the same timing convention. Identity maps induce identities, and (g∘f)_*=g_*∘f_*. These are functoriality statements proved directly from composition.

For a path c from x₀ to x₁, define a basepoint-change map from π₁(X,x₁) to π₁(X,x₀) by [α]↦[cαc̄]. Cancellation of c̄c proves multiplicativity, and the reversed path supplies the inverse. Different connecting paths can give different isomorphisms. Their difference is an inner conjugation, so a path-connected space has a fundamental group determined up to isomorphism, but not generally a canonical basepoint-free identification.

Suppose H deforms f to g and c(s)=H(x₀,s). Apply H to a loop α. The resulting parameter square has bottom f∘α, top g∘α, and both vertical sides c. Traversing its boundary gives f∘α followed by c, then the reverse of g∘α and reverse c. Since the square fills that boundary, its class is trivial. Rearranging gives f_*[α]=[c(g∘α)c̄]. If H fixes the basepoint, c is constant and the induced homomorphisms agree without conjugation.

The square-boundary argument can be formalized by contracting the boundary loop inside the square and composing with H. It does not rely on a picture alone, and its orientation determines which conjugating path appears.

Apply

If a space A is a retract of X, there are maps i:A→X and r:X→A with r∘i=id_A. Functoriality gives r_*i_*=id. Therefore i_* is injective and r_* is surjective. Later this simple algebraic consequence will rule out a disk retraction onto its boundary because the relevant groups cannot satisfy it.

Continue

Before comparing induced maps, record their source and target basepoints and any connecting path. A free homotopy generally gives conjugate data, whereas a based homotopy gives equality. The next lesson uses homotopy equivalences and deformation retracts to replace complicated spaces by simpler loop models.

Continue to Deformation retracts and homotopy equivalence.