For a universal cover \(p: \tilde{X} \to X\) there are two actions of \(\pi_1(X, x_0)\) on the fiber \(p^{-1}(x_0)\). The first is the action defined on page 69 in which the element of \(\pi_1(X, x_0)\) determined by a loop \(\gamma\) sends \(\tilde{\gamma}(1)\) to \(\tilde{\gamma}(0)\) for each lift \(\tilde{\gamma}\) of \(\gamma\) to \(\tilde{X}\), and the second is the action given by restricting deck transformations to the fiber (see Proposition 1.39). Show that these two actions are different when \(X = S^1 \vee S^1\) and when \(X = S^1 \times S^1\), and determine when the two actions are the same. [This is a revised version of the original form of this exercise.]
Solution
Fix a point \(\tilde x_0\in p^{-1}(x_0)\) and write \[ G=\pi_1(X,x_0). \] Since \(p\) is universal, every point of the fiber is the endpoint of a unique homotopy class of lifted based loops. Thus identify \[ p^{-1}(x_0)\cong G \] by sending \(h\in G\) to the endpoint of the lift of a representative loop for \(h\) starting at \(\tilde x_0\).
1. Under this identification, the usual endpoint monodromy for \(g\in G\) is right multiplication by \(g\):\[ R_g(h)=hg. \]
Start at the fiber point represented by \(h\). A path from \(\tilde x_0\) to this point projects to a loop representing \(h\). Lifting a loop representing \(g\) from this point concatenates the projected paths, so the endpoint is represented by \[ hg. \]
2. Hatcher’s first action in the exercise is the inverse of endpoint monodromy, hence\[ A_g(h)=hg^{-1}. \]
1.Therefore\[ A_g=R_g^{-1}=R_{g^{-1}}. \]
3. Under the same identification, the deck-transformation action is left multiplication:\[ D_g(h)=gh. \]
In the path-class model of the universal cover, a point is represented by a path \(h\) beginning at \(x_0\). For a loop \(g\) at \(x_0\), define \[ T_g([h])=[g h]. \] This preserves endpoints in \(X\), hence is a deck transformation. It sends the basepoint class of the constant path to the fiber point represented by \(g\), so under the standard identification \[ G\cong\operatorname{Deck}(\widetilde X/X) \] it is the deck transformation corresponding to \(g\). Restricting to the fiber gives \[ D_g(h)=gh. \]
4. The two actions agree if and only if\[ gh=hg^{-1} \qquad\text{for all }g,h\in G. \]
By 2 and 3, equality of the two actions means \[ D_g(h)=A_g(h) \] for all \(g,h\), which is exactly the displayed identity.
5. If the two actions agree, every element of \(G\) has order at most \(2\) and \(G\) is abelian.
4.Then\[ g=g^{-1}, \]so\[ g^2=e \]for every \(g\in G\).Substituting \(g^{-1}=g\) back into 4 gives\[ gh=hg \]for all \(g,h\).Thus \(G\) is abelian and every element is self-inverse.
6. Conversely, if \(G\) is abelian and every element satisfies \(g^2=e\), then the two actions agree.
For all \(g,h\), \[ A_g(h)=hg^{-1}=hg=gh=D_g(h). \]
7. Therefore the two actions are the same exactly when\[ \boxed{\pi_1(X,x_0)\text{ is an elementary abelian }2\text{-group}.} \]
This is the equivalence of 5 and 6.
8. For\[ X=S^1\vee S^1, \]the actions are different.
7.For example, at the identity fiber point,\[ A_a(e)=a^{-1}\ne a=D_a(e). \]
9. For\[ X=S^1\times S^1, \]the actions are also different.
Here \[ G\cong\mathbb Z^2. \] Although this group is abelian, its nonzero elements are not self-inverse. For a standard generator \(a\), \[ A_a(e)=a^{-1}\ne a=D_a(e). \] Thus abelianness alone does not make the actions agree.