Covering Transformations

# Covering Transformations

Definition: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. A Covering Transformation of $\tilde{X}$ is a homeomorphism $f : \tilde{X} \to \tilde{X}$ such that the following diagram commutes:
The Set of all Covering Transformations, $f : \tilde{X} \to \tilde{X}$ is denoted by $A(\tilde{X}, p)$ or simply $A(\tilde{X})$. |

The following proposition tells us that the set of all covering transformations of $\tilde{X}$ forms a group.

Proposition 1: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. Then the set of all covering transformations, $A(\tilde{X})$, forms a group with the operation of function composition. |

**Proof:**Let $f, g \in A(\tilde{X})$. Then $f \circ g$ and $g \circ f$ are also covering transformations of $\tilde{X}$ since:

\begin{align} \quad (f \circ g) \circ p = f \circ (g \circ p) = f \circ p = p \end{align}

(2)
\begin{align} \quad (g \circ f) \circ p = g \circ (f \circ p) = g \circ p = p \end{align}

- The operation $\cdot$ is associative.

- The identity function $\mathrm{id}_{\tilde{X}}$ is the identity element.

- And if $f \in A(\tilde{X})$ then $f^{-1} : \tilde{X} \to \tilde{X}$ is a homeomorphism and is such that $f^{-1} \circ p = p$, so $f^{-1} \in A(\tilde{X})$.

- Therefore $A(\tilde{X})$ is a group. $\blacksquare$

Proposition 2: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. If $f \in A(\tilde{X})$ and if for some $x_0 \in X$ we have that $f(x_0) = x_0$ then $f = \mathrm{id}_{\tilde{X}}$. That is, no covering transformation apart from the identity has fixed points. |

**Proof:**Let $f$ be a covering transformation on $\tilde{X}$ and suppose that $f(x_0) = x_0$. Then $f$ is the unique lift of $p$ starting at $x_0$. But $\mathrm{id}_{\tilde{X}}$ is also the unique lift of $p$ starting at $x_0$. Therefore:

\begin{align} \quad f = \mathrm{id}_{\tilde{X}} \end{align}

Proposition 3: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. Then for every $x \in X$ and for every $\tilde{x_1}, \tilde{x_2} \in p^{-1}(x)$ we have that $p_*(\pi_1(\tilde{X}, \tilde{x_1}) = p_*(\pi_1(\tilde{X}, \tilde{x_2}))$ if and only if there exists a unique covering transformation $f$ such that $f(\tilde{x_1}) = \tilde{x_2}$. |

**Proof:**$\Rightarrow$ Suppose that for every $x \in X$ and for every $\tilde{x_1}, \tilde{x_2} \in p^{-1}(x)$ we have that:

\begin{align} \quad p_*(\pi_1(\tilde{X}, \tilde{x_1})) = p_*(\pi_1(\tilde{X}, \tilde{x_2})) \end{align}

- Then there exists a unique lift $f : \tilde{X} \to \tilde{X}$ such that $f(\tilde{x_1}) = \tilde{x_2}$ and there exists a unique lift $g : \tilde{X} \to \tilde{X}$ such that $g(\tilde{x_2}) = \tilde{x_1}$. Then $(g \circ f)(\tilde{x_1}) = \tilde{x_1}$ and $(f \circ g)(\tilde{x_2}) = \tilde{x_2}$. By Proposition 2 this means that $f \circ g = \mathrm{id}_{\tilde{X}} = g \circ f$. So $f \circ g, g \circ f \in A(\tilde{X})$, and $f \in A(\tilde{X})$ has the property that:

\begin{align} \quad f(\tilde{x_1}) = \tilde{x_2} \end{align}

- $\Leftarrow$ Suppose that there exists a covering transformation $f \in A(\tilde{X})$ such that $f(\tilde{x_1}) = \tilde{x_2}$. Then $f$ is a lift such that $f(\tilde{x_1}) = \tilde{x_2}$. Therefore:

\begin{align} \quad p_*(\pi_1(\tilde{X}, \tilde{x_1})) \subseteq p_*(\pi_1(\tilde{X}, \tilde{x_2})) \quad (*) \end{align}

- Furthermore, since $f$ is a homeomorphism, $f$ is bijective and so $f^{-1} \in A(\tilde{X})$ is such that $f^{-1}(x_2) = x_1$. So $f^{-1}$ is a lift such that $f^{-1}(\tilde{x_2}) = \tilde{x_1}$. Therefore:

\begin{align} \quad p_*(\pi_1(\tilde{X}, \tilde{x_2}) \subseteq p_*(\pi_1(\tilde{X}, \tilde{x_1})) \quad (**) \end{align}

- From $(*)$ and $(**)$ we have that:

\begin{align} \quad p_*(\pi_1(\tilde{X}, \tilde{x_1})) = p_*(\pi_1(\tilde{X}, \tilde{x_2}) \quad \blacksquare \end{align}