Fundamental group homomorphism induced by a continuous map

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Important page, although the proof is done

Definition

Let (X,J) and (Y,K) be topological spaces[Note 1], let pX be some fixed point (to act as the base point for the fundamental group, π1(X,p)) and let φ:XY be a continuous map. Then[1]:

  • φ induces a group homomorphism on the fundamental groups, π1(X,p) to π1(Y,φ(p)), which we denote:
    • φ:π1(X,p)π1(Y,φ(p)) defined as:
    • φ:[f][φf][Note 2]

For more details on the formalities of the definition see the proof: the proof, which names everything involved during the statement.

Immediate results

The induced homomorphism of a composition is the same as the composition of induced homomorphisms

Let (X,J), (Y,K) and (Z,H) be topological spaces, let pX be any fixed point (to act as a base point for the fundamental group π1(X,p)) and let φ:XY and ψ:YZ be continuous maps. Then[1]:

  • (ψφ)=(ψφ)
    • where φ denotes the fundamental group homomorphism, φ:π1(X,p)π1(Y,φ(p)), induced by φ - and "" for the others

Note that both of these maps have the form (:π1(X,p)π1(Z,ψ(φ(p)))

The induced homomorphism of the identity map is the identity map of the fundamental group

Let (X,J) be a topological space, let IdX:XX be the identity map, given by IdX:xx and let pX be given (this will be the basepoint of π1(X,p)) then[1]:

  • the induced map on the fundamental group π1(X,p) is equal to the identity map on π1(X,p)
    • That is to say (IdX)=Idπ1(X,p):π1(X,p)π1(X,p) where Idπ1(X,p) is given by Idπ1(X,p):[f][f]

Homeomorphic topological spaces have isomorphic fundamental groups

Let (X,J) and (Y,K) be homeomorphic topological spaces, let pX be given (this will be the base point of the fundamental group π1(X,p)) and let φ:XY be that homeomorphism. Then: [1]:

That is to say:

  • (XφY)(π1(X,p)φπ1(Y,φ(p)))

Proof

See also

Notes

  1. Jump up Of course (and as usual) there is no reason why these cannot be the same spaces
  2. Jump up Caveat:Remember that [f] means f represents the equivalence class - this definition must be "well-defined" for whichever f we use. As such we must check that for απ1(X,p) that for any f and g such that f,gα that [φf]=[φg]

References

  1. Jump up to: 1.0 1.1 1.2 1.3 Introduction to Topological Manifolds - John M. Lee