In mathematics, especially in the area of topology known as algebraic topology, the induced homomorphism is a group homomorphism related to the study of the fundamental group.
Contents |
Let X and Y be topological spaces; let x0 be a point of X and let y0 be a point of Y. If h is a continuous map from X to Y such that h(x0) = y0. Define a map h* from π1(X, x0) to π1(Y, y0) by composing a loop in π1(X, x0) with h to get a loop in π1(Y, y0). Then h* is a homomorphism between fundamental groups known as the homomorphism induced by h.
h*(f 0 g) = h*(f(2t)) for t in [0,1/2] = (h*(f)) + (h*(g))
h*(f 0 g) = h*(g(2t-1)) for t in [1/2,1] = (h*(f)) + (h*(g))
so that h* is indeed a homomorphism.
Suppose X and Y are two homeomorphic topological spaces. If h is a homeomorphism from X to Y, then the induced homomorphism, h* is an isomorphism between fundamental groups [where the fundamental groups are π1(X, x0) and π1(Y, y0) with h(x0) = y0]
It has already been checked in note 2 that h* is a homomorphism. It remains to check that h* is bijective. If p is the inverse of h; then p* is the inverse of h*. This follows from the fact that (p(h))*(ƒ) = p*(h*(ƒ)) = ƒ = (h(p))*(ƒ) = h*(p*(ƒ)). If ƒ and g are two loops in X where ƒ is not homotopic to g, the h*(ƒ) is not homotopic to h*(g); if F is a homotopy between them, p*(F) would be a homotopy between ƒ and g. If k is any loop in π1(Y, y0) , then h*(p*(k)) = k where p*(k) is a loop in X. This shows that h* is bijective.
1. The torus is not homeomorphic to R2 for their fundamental groups are not isomorphic (their fundamental groups don’t have the same cardinality). A simply connected space cannot be homeomorphic to a non-simply connected space; one has a trivial fundamental group and the other does not.
2. Any two topological spaces have homomorphic fundamental groups (at a particular base point). See note 2 where h* is the homomorphism induced by the constant map. However, they need not have isomorphic fundamental groups (at a particular base point). This shows that the fundamental groups of any two topological spaces always have the same ‘group structure’.
3. The fundamental group of the unit circle is isomorphic to the group of integers. Therefore, the one-point compactification of R has a fundamental group isomorphic to the group of integers (since the one-point compactification of R is homeomorphic to the unit circle). This also shows that the one-point compactification of a simply connected space need not be simply connected.
4. The converse of the theorem need not hold. For example, R2 and R3 have isomorphic fundamental groups but are still not homeomorphic. Their fundamental groups are isomorphic because each space is simply connected. However, the two spaces cannot be homeomorphic because deleting a point from R2 leaves a non-simply connected space but deleting a point from R3 leaves a simply connected space (If we delete a line lying in R3, the space wouldn’t be simply connected anymore. In fact this generalizes to Rn whereby deleting a (n − 2)-dimensional parallelepiped from Rn leaves a non-simply connected space).
5. If A is a strong deformation retract of a topological space X, then the inclusion map from A to X yields an isomorphism between fundamental groups.