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In this thread, John Hughes gave a construction of $\pi_1$ acting on the universal cover of $X$. Let me rephrase it in the following way:

Let $p:(E,e_0) \to (X,x_0)$ be a universal covering map. Let $x_0\in X$. For any point $a$ of the covering space, there's a unique (up to homotopy relative to endpoints since $E$ is simply connected) path $\alpha$ from $e_0$ to $a$, which leads to a path $\beta = p \circ \alpha$ in $X$, from $x_0$ to $p(a)$.

Let $\gamma$ be a fixed loop in $(X,x_0)$ and $\zeta := \gamma \cdot \beta$, a path in $X$ from $x_0$ to $p(a)$. Now $\zeta$ has a unique lift to $\tilde{\zeta}$, a path in $E$ starting at $e_0$. Then the map $d_\gamma: E\to E, a\mapsto \tilde{\zeta}(1)$ is the action as desired.

I wonder if the map $d_\gamma: E\to E, a\mapsto \tilde{\zeta}(1)$ is a homeomorphism(if so, then it is a deck transformation). In particular, i don't know why $d_\gamma$ is continuous and $d_{\gamma*\gamma'}=d_{\gamma'} \circ d_\gamma $.

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so let me recap, given the universal covering $p\colon (\widetilde{X},\widetilde{x}_0) \to (X,x_0)$ we say that:

Def A deck transformation $\tau \colon \widetilde{X}\to \widetilde{X}$ is a lift of the covering map $p$

continuity of a deck transformation is then clear almost by definition: by Prop 1.33 on Hatcher's book (page 61) (check continuity locally using the fact that $p$ restricts to a local homeomorphism).

After looking carefully at the construction of the lift of $p$ (Prop 1.33 on Hatcher's book) one sees that $$\tau(y):= \widetilde{p\gamma}(1)$$ where $\gamma$ is a path in $\widetilde{X}$ s.t. $\gamma(0)=\widetilde{x}_0$, $\gamma(1)=y$. Let us set $\widetilde{p\gamma}(1)=z$.

The choice of such path is in influent, thanks to the simply-connectedness of $\widetilde{X}$ any two of such choice are homotopic rel. end-points and by the homotpy lifting property the end points of the lifts are preserved (See again prop. 1.33).

Now we construct $\tau^{-1}$ similarly, but using this time the inverse paths (w.r.t. to the choices done for constructing $\tau$). In particular, $$\tau^{-1}(z):=\widetilde{p\overline{\gamma}(1)}$$ but by uniqueness of lifts, the lift of the inverse path $\overline{\gamma}$ from the point $z$ is the path $\overline{\widetilde{p\gamma}}$, so by def of inverse path $$\tau^{-1}(z):=\widetilde{p\overline{\gamma}(1)}= \widetilde{p\gamma}(0)=\widetilde{x}_0$$

Since we just showed that $\tau^{-1}\tau(\widetilde{x}_0)=\widetilde{x}_0$, by the uniqueness of lifts (the composition of two deck transformations is again a deck transformation, I leave this easy proof to the reader) applied to the map and the identity $Id_{\widetilde{X}}$ (which is always a lift of $p$) we get that $$\tau^{-1}\tau=Id$$ The other composition is done in the same way.

So back to the original question: how do we associate an element of $\pi_1(X,x_0)$ to a Deck transformation? Well, in the obvious way: $$\alpha \in \pi_1(X,x_0) \mapsto \tau_{\alpha}$$ where $\tau_{\alpha}$ is the unique deck transformation which sends $\tilde{x}_0$ to $\widetilde{\alpha}_{\widetilde{x}_0}(1)$, for $\widetilde{\alpha}_{\widetilde{x}_0}$ the unique lift starting at $\tilde{x}_0$. The fact that this association is a morphism should be clear now: again you want to use the uniqueness of liftings to argue that $$\tau_{\alpha}\tau_{\beta}=\tau_{\beta*\alpha}$$ (for me concatenation of path goes from left to right). In fact it's enough to prove that the two deck transformations coincide on one point: $$\tau_{\alpha}\tau_{\beta}(\widetilde{x}_0)=\tau_{\alpha}\left(\widetilde{\beta}_{\widetilde{x}_0}(1) \right)\overset{def}{=}\widetilde{\alpha}_{\widetilde{\beta}_{\widetilde{x}_0}(1)}(1)$$ $$\tau_{\beta*\alpha}(\widetilde{x}_0)=\widetilde{\beta *\alpha}_{\widetilde{x_0}}(1)$$ But the two results are the same again by the uniqueness of lifts. Since you build the lift piece by piece, when you lift $\beta *\alpha$ and you are done lifting the first piece (i.e. $\beta$), you find yourself starting lifting $\alpha$ starting from $\widetilde{\beta}(1)$, i.e. you consider the unique lift $\widetilde{\alpha}_{\widetilde{\beta}_{\widetilde{x}_0}(1)}$, which concludes the proof.

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