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

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Statement

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:

• 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]

Proof

The proof is simple.

• Let [ilmath][f]\in\pi_1(X,p)[/ilmath] be given.
  • By the fundamental group homomorphism induced by a continuous map we can deal with [ilmath]f[/ilmath] and the induced morphism is well defined.
  • [ilmath](\text{Id}_X)_*([f])[/ilmath]

    [ilmath]\eq[\text{Id}_X\circ f][/ilmath]

    [ilmath]\eq[f][/ilmath]

• Since [ilmath][f]\in\pi_1(X,p)[/ilmath] was arbitrary we have shown it for all.

This is the very definition of [ilmath](\text{Id}_X)_*[/ilmath] being the identity map on [ilmath]\pi_1(X,p)[/ilmath]