How to define an action of $\pi_1(E)$ on $\pi_n(F)$ for a fibration $F\to E\to B$?

Question

Given a fibration $F\to E\to B$, how to show that the action $\pi_1(F)$ on $\pi_n(F)$ can factor through $\pi_1(E)$: $\pi_1(F)\to\pi_1(E)\to\text{Aut}(\pi_n(F))$? This is exercise 4.3.10 from Hatcher's Algebraic topology.

Answer 1

Let $p:E\rightarrow B$ the projection and $i:F\hookrightarrow E$ the fibre inclusion. Choose a common basepoint $e_0\in F\subseteq E$ for both spaces and let $b_0=p(e_0)\in B$ be the basepoint of $B$.

Let $[\beta]\in\pi_1(E,e_0)$ be the homotopy class of a loop $\beta:I\rightarrow E$ and consider the following commutative diagram

$\require{AMScd}$ \begin{CD} F\times0\cup e_0\times I@>i\cup\beta>> E\\ @VV V @VV p V\\ F\times I @>p\circ\beta\circ pr_2>> B. \end{CD}

Then the homotopy lifting property of the fibration produces a map $\tilde\beta:F\times I\rightarrow E$ satisfying $\tilde\beta(f,0)=i(f)$, $\tilde\beta(e_0,t)=\beta(t)$ and $p\circ \tilde\beta(f,t)=p\circ \beta(t)$. We note that $p\circ\tilde\beta(f,1)=p\circ \beta(1)=b_0$, so for each $f\in F$, the point $\tilde\beta(f,1)$ actually lies in the fibre $F$. We set $\tilde\beta_1=\tilde\beta(-,1):F\rightarrow F$.

One shows by standard arguments that the homotopy class of $\tilde\beta_1$ depends only on that of $\beta$, and is uniquely defined by it. The loop $\beta$ has an inverse $\beta^{-1}$ in $\pi_1(E,e_0)$, and from this it follows that $\tilde\beta_1$ has a homotopy inverse $\widetilde{\beta^{-1}}_1$, and so is a homotopy equivalence.

Thus we have a map

$$\pi_1(E,e_0)\rightarrow \pi_0Aut_*(F),\qquad \beta\mapsto\tilde\beta_1.$$

This gives an action of $\pi_1(E,e_0)$ on $\pi_n(F,e_0)$ by declaring

$$\beta\cdot \alpha=\tilde\beta_1\circ\alpha$$

for $\beta\in\pi_1(E,e_0)$, $\alpha\in\pi_n(F,e_0)$.

Now the action of $\pi_1(F,e_0)$ on $\pi_n(F,e_0)$ may be obtained similarly. In particular by applying exactly the same procedure to the fibration $F\rightarrow\ast$.

Now let $\alpha\in\pi_1(F,e_0)$ and consider the following diagram

$\require{AMScd}$ \begin{CD} F\times0\cup e_0\times I@>id_F\cup\alpha>> F@>i>>E\\ @V V V @VV V@VV p V\\ F\times I @>>> \ast@>>>B \end{CD}

which commutes strictly since $p\circ i=\ast$. Applying the homotopy lifting property to the left-hand square gives the map $\tilde\alpha:F\times I\rightarrow F$ which defines $\tilde\alpha_1:F\rightarrow F$ and so specifies the action of $\alpha\in\pi_1(F,e_0)$ on $\pi_n(F,e_0)$.

On the other hand, if we apply the homotopy lifting property to the combined square we get a map $\widetilde{i\alpha}:F\times I\rightarrow E$ which defining $\widetilde{i\alpha}_1:F\rightarrow F$ andspecifying the action of $i\circ\alpha\in\pi_1(E,e_0)$ on $\pi_n(F,e_0)$.

Now the arguments for uniqueness of the resulting maps apply, so since the diagram commutes strictly we may take $\widetilde{i\alpha}=i\circ\widetilde\alpha:F\times I\rightarrow E$. This tells us that $\widetilde{i\alpha}(-,1)\simeq i\circ\tilde\alpha(-,1):F\rightarrow E$, so that in particular

$$\widetilde{i\alpha}_1=\tilde\alpha_1\in\pi_0Aut_*(F).$$

Thus if $\gamma\in\pi_n(F,e_0)$ we have

$$(i\alpha)\cdot\gamma=\widetilde{i\alpha}_1\circ\gamma\simeq \tilde\alpha_1\circ\gamma=\tilde\alpha_1\cdot\gamma$$

where the left hand side is the action of $\alpha=i_*\alpha\in\pi_1(E,e_0)$ on $\gamma$, and the right hand side is the action of $\alpha\in\pi_1(F,e_0)$ on $\gamma$.

We're done.