6

Let $p:(E, e_0)\to (X, x_0)$ be a covering. If $\alpha:[0, 1]\to X$ is a path beggining at $x_0$ and $e\in p^{-1}(x_0)$, then we can uniquely lift $\alpha$ to a path $\alpha'_e:[0, 1]\to E$ beggining at $e$, such that $p\alpha'_e = \alpha$.

Moreover, if $\alpha$ is homotopic to $\beta$ relative to $\{0, 1\}$, then $\alpha'_e$ is homotopic to $\beta_e'$ relative to $\{0, 1\}$. In particular, $\alpha'_e$ and $\beta'_e$ have the same endpoints.

If $e\in p^{-1}(x_0)$ and $[\sigma]\in \pi_1(X, x_0)$, then we define function $\cdot:p^{-1}(x_0)\times\pi_1(X, x_0) \to p^{-1}(x_0)$ as $e\cdot [\sigma] = \sigma'_e(1) \in p^{-1}(x_0)$.

We want to prove that $\cdot$ is an action of $\pi_1(X, x_0)$ on the fiber $p^{-1}(x_0)$.

Constant loop on $x_0$ will be denoted by the same symbol etc.

If $e$ is such loop, then $pe = x_0$, so that $e$ is the lift of $x_0$ beggining at $e$. Then $e\cdot [x_0] = e$.

Now suppose that $[\sigma], [\tau]\in\pi_1(X, x_0)$ and $e\in p^{-1}(x_0)$. Then $e\cdot ([\sigma]\cdot [\tau]) = (\sigma\tau)'_e(1)$ and if we denote $\sigma_e'(1) = e_1$, then $(e\cdot [\sigma])\cdot [\tau] = \tau'_{e_1}(1) = \sigma_e'\tau_{e_1}'(1)$. Now, it suffices to prove that $\sigma'$ and $\tau'$ are two paths such that $\sigma'(1) = \tau'(0)$, then $p(\sigma'\tau') = p(\sigma')p(\tau')$.

When searching for a proof that $\cdot$ is an action on the fiber $p^{-1}(x_0)$ in books, most just mentioned that it is, and the proof I found in Rotman's book left at saying that $p(\sigma'\tau') = p(\sigma')p(\tau')$. I was given this as an exercise, and I could tell you have to prove that $p(\sigma'\tau'), p(\sigma')p(\tau')$ are at least homotopic, but it's not obvious to me why it would be. Another proof I found on one of the videos on algebraic topology course, started with showing that those are permutations, but still I found a gap in authors logic, since they assumed inverse of lift of $\alpha$ is the same as lift of inverse of $\alpha$.

From here comes my question, why is $\cdot$ an action, or more precisely, why $p(\sigma'\tau') = p(\sigma')p(\tau')$ (where $\sigma'$, $\tau'$ are in the same setting as above).

Jakobian
  • 10,247

1 Answers1

2

So $e \cdot [c]$ is $\overline{c}_e(1)$, where $\overline{c}_e$ is the lift of $c$ starting at $e$. (I will use a line instead of $'$ to indicate the lift.)

By definition, $e \cdot ([c]*[d])=e\cdot([c*d])=\overline{(c*d)}_e(1)$. Note that $\overline{c *d}_e=\overline{c}_{\overline{d}_e(1)}*\overline{d}_{e}$. This is true simply because $p \circ (\overline{c}_{\overline{d}(1)} * \overline{d}_{e})=c*d$ $^{(1)}$and lifts of paths with prescribed initial points are unique.

Thus, $$(e\cdot [d]) \cdot [c]=\overline{c}_{(e \cdot [d])}(1)=\overline{c}_{\overline{d}_e(1)}(1)=(\overline{c}_{\overline{d}_e(1)}*\overline{d}_{e})(1)=\overline{c *d}_e(1)=e \cdot ([c]*[d]).$$

$^{(1)}$ Maybe it is here that your core doubt lies, so let's make this explicit:

For $0 \leq t \leq 1/2$, \begin{align*} p \circ (\overline{c}_{\overline{d}(1)} * \overline{d}_{e})(t) &=p( (\overline{c}_{\overline{d}(1)} * \overline{d}_{e})(t)) \\ &=p(\overline{d}_e(2t)) \\ &=d(2t) \\ &=c*d(t). \end{align*} For $1/2 \leq t \leq 1$, \begin{align*} p \circ (\overline{c}_{\overline{d}(1)} * \overline{d}_{e})(t) &=p( (\overline{c}_{\overline{d}(1)} * \overline{d}_{e})(t)) \\ &=p(\overline{c}_{\overline{d}(1)}(2t-1)) \\ &=c(2t-1) \\ &=c*d(t). \end{align*}

Aloizio Macedo
  • 34,292