2
$\begingroup$

Any help on the following question will be great.

Question: Given a map $f: (X,x) \rightarrow (Y,y)$ of connected and locally simply connected spaces, show that it induces a unique map of universal covering spaces $\hat{f}: (\hat{X},\hat{x}) \rightarrow (\hat{Y},\hat{y})$ such that the diagram below commutes.

$$\begin{equation*}\begin{array}{ccl} (\hat{X},\hat{x})&\stackrel{\hat{f}}{\longrightarrow}& (\hat{Y},\hat{y})\\ \!\!\!\!\!\!\!\!{\scriptstyle p}\downarrow& &\downarrow{\scriptstyle q}\\ \!\!\!\!(X,x) &\stackrel{f}{\longrightarrow}& (Y,y) \end{array}\qquad\qquad\end{equation*}$$

Thanks.

$\endgroup$
  • 2
    $\begingroup$ You get a map $f \circ p: \hat{X} \rightarrow Y$. Then examine the lifting criteria in terms of $\pi_1$. $\endgroup$ – Joe Johnson 126 Mar 28 '17 at 12:48
  • 1
    $\begingroup$ You should have a fact that a map $g: Z \to Y$ induces a map $\hat{g}: Z \to \hat{Y}$ when a certain condition is met for the induced map $g_*: \pi_1(Z) \to \pi_1(Y)$. $\endgroup$ – Joe Johnson 126 Mar 28 '17 at 15:28
2
+50
$\begingroup$

Existence: For every $\hat{z} \in \hat{X}$ pick a curve $\hat{\gamma}$ in $\hat{X}$ from $\hat{x}$ to $\hat{z}$, that is $\hat{\gamma} (0) = \hat{x}$, and $\hat{\gamma} (1) = \hat{z}$, and let $\gamma = p \circ \hat{\gamma}$. Let $\hat{\Gamma}$ be the unique lift of $f \circ \gamma$ in $Y$ to $\hat{Y}$ starting at $\hat{y}$. Define $\hat{f}(\hat{z}) = \hat{\Gamma} (1)$. It's easy to see, that this does not depend on the choice of $\hat{\gamma}$, because if $\tilde{\gamma}$ is another such curve, then they are homotopic (while the endpoints are fixed), because $\hat{X}$ is simply-connected, thus $f \circ p \circ \hat{\gamma}$ is homotopic to $f \circ p \circ \tilde{\gamma}$ (while the endpoints are fixed). Thus their unique lifts to $\hat{Y}$ starting at $\hat{y}$ has the same endpoints too (due to the properties of the lifts).

Uniqueness: Let $\hat{z} \in \hat{X}$, and $\hat{f}'$ another function satisfying the conditions. Pick again any curve $\hat{\gamma}'$ in $\hat{X}$ from $\hat{x}$ to $\hat{z}$. Now $\hat{f}' \circ \hat{\gamma}' = \hat{\Gamma}'$ is a curve like $\hat{\Gamma}$ was in the Existence part, thus $\hat{f}' (\hat{z}) = \hat{f}' (\hat{\gamma}'(1)) = \hat{\Gamma}' (1) = \hat{f} (\hat{z})$, thus $\hat{f}' = \hat{f}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.