The Uniqueness of Universal Covers Theorem

The Uniqueness of Universal Covers Theorem

Recall from the Lifts of Paths page that if $X$ is a topological space, $(\tilde{X}, p)$ is a covering space of $X$, $Y$ is a path connected and locally path connected topological space, and $f : Y \to X$ is a continuous function with $y_0 \in Y$, $x_0 = f(y_0)$, and $\tilde{x_0} \in p^{-1}(x_0)$ then there exists a unique lift of $f$, call it $\tilde{f} : Y \to \tilde{X}$ if and only if:

(1)
\begin{align} \quad f_*(\pi_1(Y, y_0)) \subseteq p_*(\pi_1(\tilde{X}, \tilde{x_0})) \end{align}
Theorem 1 (The Uniqueness of Universal Covers Theorem): Let $X$ be a topological space and suppose that $(\tilde{X_1}, p_1)$ and $(\tilde{X_2}, p_2)$ are both universal covers of $X$. Then $\tilde{X_1}$ is homeomorphic to $\tilde{X_2}$.
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  • Proof: Let $x \in X$. Take $\tilde{x_1} \in p_1^{-1}(x)$ and take $\tilde{x_2} \in p_2^{-1}(x)$. Since $(\tilde{X_1}, p_1)$ and $(\tilde{X_2}, p_2)$ are universal covers of $X$ we have that $\pi_(\tilde{X_1}, \tilde{x_1})$ and $\pi_1(\tilde{X_2}, \tilde{x_2})$ are both the trivial group, and so:
(2)
\begin{align} \quad p_{1*}(\pi_(\tilde{X_1}, \tilde{x_1})) &= \{ 1 \} \\ \quad p_{2*}(\pi_1(\tilde{X_2}, \tilde{x_2})) &= \{ 1 \} \end{align}
  • Therefore any subset of these spaces is a subgroup. Hence there exists unique lifts $f : \tilde{X_1} \to \tilde{X_2}$ and $g : \tilde{X_2} \to \tilde{X_1}$ such that $f(x_1) = x_2$ and $g(x_2) = x_1$.
  • Now observe that $g \circ f : \tilde{X_1} \to \tilde{X_1}$ is also a lift of $p_1$ and:
(3)
\begin{align} \quad (g \circ f)(x_1) = g(f(x_1)) = g(x_2) = x_1 \end{align}
  • By the uniqueness of lifts we must have that $g \circ f = \mathrm{id}_{\tilde{X_1}}$.
  • Similarly, observe that $f \circ g : \tilde{X_2} \to \tilde{X_2}$ is also a lift of $p_2$ and:
(4)
\begin{align} \quad (f \circ g)(x_2) = f(g(x_2)) = f(x_1) = x_2 \end{align}
  • By the uniqueness of lifts we must have that $f \circ g = \mathrm{id}_{\tilde{X_2}}$.
  • Hence both $f$ and $g$ are homeomorphisms, so $\tilde{X_1}$ and $\tilde{X_2}$ are homeomorphic. So universal covers are unique up to homeomorphisms. $\blacksquare$
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