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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.

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    $\begingroup$ You get a map $f \circ p: \hat{X} \rightarrow Y$. Then examine the lifting criteria in terms of $\pi_1$. $\endgroup$
    – J126
    Commented Mar 28, 2017 at 12:48
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    $\begingroup$ 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 $g_*: \pi_1(Z) \to \pi_1(Y)$. $\endgroup$
    – J126
    Commented Mar 28, 2017 at 15:28

1 Answer 1

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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 $\hat{f}' = \hat{f}$.

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