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In some lecture notes I am reading through there is this result and a proof of it that I don't seem to understand.

Lemma: If $F:X→X$ is a continuous map which is homotopic to the identity map $id_X:X→X$, then the induced map $F_∗:π_1(X,x)→π_1(X,F(x))$ is an isomorphism.

Proof: Let $F_t:X→X$ be a homotopy from $F_0=F$ to $F_1=id_X$ and let $δ$ be the path traced out by the basepoint x under this homotopy, that is $$δ(t)=F_t(x)$$ Now, for $[γ]∈π_1(X,x)$, we have $$F_∗[γ]=[F∘γ]$$ and $F∘γ$ is freely homotopic to $γ$ via the free homotopy $F_s∘γ$. This is a free (rather than based) homotopy because the basepoint of $F_s∘γ$ is $δ(s)$. Using our results on basepoint dependence, this implies that $$[F∘γ]=[δ⋅γ⋅δ^{−1}]$$ and that the map $[γ]→[δ⋅γ⋅δ^{−1}]$ is an isomorphism $π_1(X,x)→π_1(X,F(x))$. Since $F_∗[γ]=[F∘γ]$, this tells us that $F_∗$ is an isomorphism. $\square$

I don't understand how $[F∘γ]=[δ⋅γ⋅δ^{−1}]$ is a consequence of 'the results on basepoint dependence', which are:

  1. Given a path $δ:[0,1]→X$ with $δ(0)=x$ and $δ(1)=y$ we obtain an isomorphism $F_δ:π_1(X,x)→π_1(X,y)$ given by $F_δ([γ])=[δ⋅γ⋅δ^{−1}]$

  2. Free homotopy classes of loops based at $x$ are conjugacy classes in $π_1(X,x)$

Could someone kindly elaborate on that?

Mafematician
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1 Answers1

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It doesn't really seem like they use the dependence of basepoint results directly, but more like they use a technique that's similar to the one used in the proof of these dependence results.

To show that $[F \circ \gamma] = [\delta \cdot \gamma \cdot \delta^{-1}]$ directly, you need to connect the two loops using a continuous family of loops based at $F(x)$. The canonical choice is $\delta_t \cdot (F_t \circ \gamma) \cdot \delta_t^{-1}$, where $\delta_t$ is the path from $F(x)$ to $F_t(x)$. You can visualise this nicely by drawing a picture.

Motmot
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