Topology/Induced homomorphism

The induced homomorphism is related to the study of the fundamental group. We will give a few theorems and notes, but first we make a definition.

Definition

Let X and Y be topological spaces; let x0 be a point of X and let y0 be a point of y. Suppose h is a continuous map from X to Y such that h(x0) = y0. Define a map h* from π1(X,x0) to π2(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.

Note 1

Let us check that if f is a loop in π1(X,x0), then h*(f) is a loop in π1(Y,y0). Note that h*(f) is a continuous map from [a,b] to Y (we will assume that [a,b] = [0,1] which has no difference to the general case since these two sets are homeomorphic), and h*(f(0)) = h*(x0) = y0 and h*(f(1)) = h*(x0) = y0.

Note 2

We will check that h is indeed a homomorphism. To avoid repetition, whenever we call f and g loops, they will be known as loops based at x0. Suppose f and g are two loops. Then [0 is the group operation on π1(X,x0) and + is the group operation on π1(Y,y0)]

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.

Note 3

Checking h* is a function (i.e, every loop in π1(X,x0) gets mapped onto a unique loop in π1(Y,y0)) follows from the fact that if f and g are loops in π1(X,x0) that are homotopic via the homotopy H, then h*(f) and h*(g) are homotopic via the homotopy h*H.

Note 4

Note that none of the above notes would be true unless h is continuous which is why this is needed in the hypothesis. We leave it to you to work out why h(x0) = y0 [which is fairly trivial].

We now prove one important theorem which can be used to check whether two topological spaces are homeomorphic or not. In fact, this theorem illustrates why algebraic topology was invented in the first place.

Theorem

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 [Assume that we are considering the fundamental groups π1(X,x0) and π1(Y,y0) with h(x0) = y0]

Proof

We have already checked in note 2 that h* is a homomorphism. It remains to check that h* is bijective. Suppose p is the inverse of h; then p* is the inverse of h*. This follows from the fact that (p(h))*(f) = p*(h*(f)) = f = (h(p))*(f) = h*(p*(f)). If f and g are two loops in X where f is not homotopic to g, the h*(f) is not homotopic to h*(g); if F is a homotopy between them, p*(F) would be a homotopy between f 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.

Note: It is a good exercise to check that we used all the properties that h satisfies, i.e we used completely, the fact that h is a homeomorphism.

Applications of the Theorems

1. The torus is not homeomorphic to R^2 for their fundamental groups are not isomorphic (their fundamental groups don’t have the same cardinality). Note that, 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. In fact, any two topological spaces have homomorphic fundamental groups (at a particular base point). See note 2 where we may let h* be the homomorphism induced by the constant map. However, they need not have isomorphic fundamental groups (at a particular base point). This is interesting because it 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 additive group of integers. Therefore, the fundamental group of [0,1] is isomorphic to the set of integers since [0,1] and the unit circle are homeomorphic (why is this statement false?). The one-point compactification of R also has a fundamental group isomorphic to the set of integers (since the one-point compactification of R is homeomorphic to the unit circle).

4. The converse of the theorem need not hold. For example, R^2 and R^3 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 R^2 leaves a non-simply connected space but deleting a point from R^3 leaves a simply connected space (If we delete a line lying in R^3, the space wouldn’t be simply connected anymore. In fact this generalizes to R^n whereby deleting a (n-2) dimensional parallelepiped from R^n leaves a non-simply connected space).