need clarification in Hatcher's Algebraic Topology Exercise 1.3.27

First, http://pi.math.cornell.edu/~hatcher/AT/AT.pdf is the site of the book.

Exercise 1.3.27 is :

Exercise 1.3.27. For a universal cover $$p : \tilde X \to X$$ we have two actions of $$\pi_1(X, x_0)$$ on the fiber $$p^{-1}(x_0)$$, namely the action given by lifting loops at $$x_0$$ and the action given by restricting deck transformations to the fiber. Are these two actions the same when $$X=S_1 \vee S_1$$ or $$X=S_1 \times S_1$$? Do the actions always agree when $$\pi_1(X,x_0)$$ is abelian?

My interpretation of the two actions are as :

(1) The first action is given in p.69 and defined as follows : Let $$[\gamma] \in \pi_1(X ,x_0)$$ and let $$\tilde x \in p^{-1}(x_0)$$. Then there is a unique lift $$\delta$$ of $$\bar{\gamma}$$ starting at $$\tilde x$$. We define $$[\gamma] \tilde x$$ to be the point $$\delta (1)$$. (Hatcher used $$\bar{\gamma}$$, not $$\gamma$$, to make the action a left action. Also, as in the definition of an action in p.71, every action in this book means a left action. )

(2) The second action is : Fix $$\tilde x _0 \in p^{-1}(x_0)$$. Let $$[\gamma] \in \pi_1(X ,x_0)$$ and let $$\tilde x \in p^{-1}(x_0)$$. Since $$p$$ is a simply-connected covering, there is a unique deck transformation $$\tau_\gamma$$ sending $$\tilde x _0$$ to $$\tilde \gamma (1)$$, where $$\tilde \gamma$$ is the lift of $$\gamma$$ starting at $$\tilde x _0$$ (This is Proposition 1.39. In fact $$\pi_1(X, x_0)$$ is isomorphic to $$G(\tilde X)$$, the group of deck transformations, via the isomorphism sending $$[\gamma]$$ to $$\tau_\gamma$$.) Now we define $$[\gamma]\tilde x$$ to be the point $$\tau_\gamma (\tilde x)$$.

Assuming my understand is correct, what I am confused is these two actions are trivially different. Am I wrong with the definitions of the actions?

• Why do you say they are trivially different? Aug 26, 2019 at 3:16
• I don't see the difference, can you explain ? Except that in 1) you are more or less constructing the universal cover as the set of curves in $X$ modulo homotopy (so $\bar{X}$ is a manifold) and in 2) you are assuming you are given a simply connected space with a covering Aug 26, 2019 at 3:25
• @EricWofsey For example, consider $X=S_1 \vee S_1$ as a graph (see p.57 of Hatcher) and the universal covering space of $S_1 \vee S_1$ given in Example 1.45(p.77). Let $\alpha \in \pi_1(X, x_0)$ be the class of the loop $a$ in $X$ (where $x_0$ is the obvious vertex in $X$). Then for the first action, $\alpha e=a^{-1}$ while $\alpha e=a$. This argument is similar for the torus, where the universal covering spcae of the torus is $\Bbb R^2$ with grids along $\Bbb Z \times \Bbb Z$ Aug 26, 2019 at 3:43

These are indeed not the same in most cases (including when $$X=S^1\vee S^1$$ or $$X=S^1\times S^1$$) for the rather trivial reason that the first action is defined using $$\bar{\gamma}$$ and the second is defined using $$\gamma$$ (so at least when acting on the basepoint $$\tilde{x}_0$$, the action of $$[\gamma]$$ by the first definition corresponds to the action of $$[\gamma]^{-1}$$ by the second). I'm guessing that what Hatcher had in mind, though, is the version of the first action which is defined using a lift of $$\gamma$$, not a lift of $$\bar{\gamma}$$. Of course, that makes the first action a right action rather than a left action, so the actions cannot possibly be the same when $$\pi_1(X,x_0)$$ is nonabelian (given that the actions are faithful). But for an abelian group, right and left actions are the same, and so this still leaves a nontrivial question of whether they coincide when $$\pi_1(X,x_0)$$ is abelian.
• How can it not be true for when $X = S^1 \times S^1$? The fundamental group is abelian. Dec 3, 2020 at 8:14
• Identifying the fiber with $\pi_1(X,x_0)$, one action has $g\in \pi_1(X,x_0)$ act by left translation by $g$ and the other has it act by right translation by $g^{-1}$. Dec 3, 2020 at 15:24