This is for homework.

I'm supposed to do exercise 4.1.4 in Hatchers "Algebraic Topology", which is to show that given a universal covering $p: \tilde{X} \to X$ of a path-connected space $X$, the action of $\pi_1(X)$ on $\pi_n(X)$ can be identified with the induced action of the group of deck transformations $G(\tilde{X}) \simeq \pi_1(X)$ (since $p$ is the universal cover) on $\pi_n(\tilde{X})$.

I think maybe my problem lies in the fact that I don't have a good intuition of the action of $\pi_1(X)$ on $\pi_n(X)$, but anyway, this is what I've done:

Let $[\gamma] \in \pi_1(X)$ be the class of the loop $\gamma: I \to X$ at the basepoint, and denote by $d_\gamma: \tilde{X} \to \tilde{X}$ the associated deck transformation. From this we can induce an automorphism $d_\gamma * : \pi_n(\tilde{X}) \to \pi_n(\tilde{X})$ given by $d_\gamma*[\tilde{f}] = [d_\gamma \circ \tilde{f}]$ for some $\tilde{f} : S^n \to \tilde{X}$.

Thus to each element $[\gamma]$ of $\pi_1(X)$ we can associate an automorphism $\pi_n(\tilde{X}) \to \pi_n(\tilde{X})$. This is the action of $\pi_1(X)$ on $\pi_n(\tilde{X})$.

Now from $p: \tilde{X} \to X$ we get isomorphisms of the higher homotopy groups of $X$ and $\tilde{X}$, e.g. $p_*: \pi_n(\tilde{X}) \to \pi_n(X)$.

From what I've understood I want to show that the induced action of a group element $[\gamma]$ on $\pi_n(\tilde{X})$ through this isomorphism is equal to the action of $[\gamma]$ on $\pi_n(X)$, where the action of $[\gamma]$ is the automorphism $[f] \mapsto [\gamma f]$ where $\gamma f: S^n \to X$ is as described in the book (it is the "canonical" action of $\pi_1(X)$ on $\pi_n(X)$).

In other words I want to show that $p_*(d_\gamma*[\tilde{f}] ) = [\gamma f]$ for $f: S^n \to X$ having $\tilde{f}: S^n \to \tilde{X}$ as lift.

But my problem is this: Since $d_\gamma$ is a deck transformation (a permutation on the fibers), I have $(p \circ d_\gamma)(\tilde{x}) = p(\tilde{x})$ for all $\tilde{x} \in \tilde{X}$ and $d_\gamma \in G(\tilde{X})$, thus I get that $$ p_* \left( d_\gamma*[\tilde{f}] \right) = p_* \left( [d_\gamma \circ \tilde{f}] \right) = [p \circ d_\gamma \circ \tilde{f}] = [p \circ \tilde{f}] = [f] $$ Thus if the statement I think I'm proving is true, then $$ [\gamma f] = p_* \left(d_\gamma*[\tilde(f)] \right) = [f] $$ for all $[\gamma] \in \pi_1(X)$ and the action is trivial.

Thanks in advance for your help!


The issue is with base points:

You write:

From this we can induce an automorphism $d_\gamma*:\pi_n(\tilde X)\rightarrow\pi_n(\tilde X)$ given by $d_\gamma *[\tilde f]=[d_\gamma\circ\tilde f]$ for some $\tilde f:S^n\rightarrow \tilde X$.

But notice that $[\tilde f]$ has a different base point than $[d_\gamma\circ\tilde f]$ so it's not quite right to say $d_\gamma *$ is an "automorphism", since technically, $[\tilde f]$ and $[d_\gamma\circ\tilde f]$ belong to different groups.

They are isomorphic groups and so it seems like the better thing to do would be what you did, then move back to the group $[\tilde f]$ lives in by the isomorphism $\tilde\gamma_ *^{-1}$ which is just a change of base point transformation in $\tilde X$ via the path $\tilde\gamma$ which is the lift of $\gamma$ that corresponds to the deck transformation $d_\gamma$.

Being a lift of $\gamma$, this should now work out the way you want it to.


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