Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.