Fundamental group homomorphism induced by a continuous map

From Maths
Jump to: navigation, search
Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Important page, although the proof is done


Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces[Note 1], let [ilmath]p\in X[/ilmath] be some fixed point (to act as the base point for the fundamental group, [ilmath]\pi_1(X,p)[/ilmath]) and let [ilmath]\varphi:X\rightarrow Y[/ilmath] be a continuous map. Then[1]:

  • [ilmath]\varphi[/ilmath] induces a group homomorphism on the fundamental groups, [ilmath]\pi_1(X,p)[/ilmath] to [ilmath]\pi_1(Y,\varphi(p))[/ilmath], which we denote:
    • [ilmath]\varphi_*:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))[/ilmath] defined as:
    • [ilmath]\varphi_*:[f]\mapsto [\varphi\circ f][/ilmath][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 [ilmath](X,\mathcal{ J })[/ilmath], [ilmath](Y,\mathcal{ K })[/ilmath] and [ilmath](Z,\mathcal{ H })[/ilmath] be topological spaces, let [ilmath]p\in X[/ilmath] be any fixed point (to act as a base point for the fundamental group [ilmath]\pi_1(X,p)[/ilmath]) and let [ilmath]\varphi:X\rightarrow Y[/ilmath] and [ilmath]\psi:Y\rightarrow Z[/ilmath] be continuous maps. Then[1]:

  • [ilmath](\psi\circ\varphi)_*\eq(\psi_*\circ\varphi_*)[/ilmath]
    • where [ilmath]\varphi_*[/ilmath] denotes the fundamental group homomorphism, [ilmath]\varphi_*:\pi_1(X,p)\rightarrow\pi_1(Y,\varphi(p))[/ilmath], induced by [ilmath]\varphi[/ilmath] - and "" for the others

Note that both of these maps have the form [ilmath]\big(:\pi_1(X,p)\rightarrow\pi_1(Z,\psi(\varphi(p))\big)[/ilmath]

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

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space, let [ilmath]\text{Id}_X:X\rightarrow X[/ilmath] be the identity map, given by [ilmath]\text{Id}_X:x\mapsto x[/ilmath] and let [ilmath]p\in X[/ilmath] be given (this will be the basepoint of [ilmath]\pi_1(X,p)[/ilmath]) then[1]:

  • the induced map on the fundamental group [ilmath]\pi_1(X,p)[/ilmath] is equal to the identity map on [ilmath]\pi_1(X,p)[/ilmath]
    • That is to say [ilmath](\text{Id}_X)_*\eq\text{Id}_{\pi_1(X,p)}:\pi_1(X,p)\rightarrow\pi_1(X,p)[/ilmath] where [ilmath]\text{Id}_{\pi_1(X,p)} [/ilmath] is given by [ilmath]\text{Id}_{\pi_1(X,p)}:[f]\mapsto [f][/ilmath]

Homeomorphic topological spaces have isomorphic fundamental groups

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be homeomorphic topological spaces, let [ilmath]p\in X[/ilmath] be given (this will be the base point of the fundamental group [ilmath]\pi_1(X,p)[/ilmath]) and let [ilmath]\varphi:X\rightarrow Y[/ilmath] be that homeomorphism. Then: [1]:

That is to say:

  • [ilmath]\big(X\cong_\varphi Y)\implies(\pi_1(X,p)\cong_{\varphi_*}\pi_1(Y,\varphi(p))\big)[/ilmath]


See also


  1. Of course (and as usual) there is no reason why these cannot be the same spaces
  2. Caveat:Remember that [ilmath][f][/ilmath] means [ilmath]f[/ilmath] represents the equivalence class - this definition must be "well-defined" for whichever [ilmath]f[/ilmath] we use. As such we must check that for [ilmath]\alpha\in\pi_1(X,p)[/ilmath] that for any [ilmath]f[/ilmath] and [ilmath]g[/ilmath] such that [ilmath]f,g\in\alpha[/ilmath] that [ilmath][\varphi\circ f]\eq[\varphi\circ g][/ilmath]


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