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I was confused by this post, and I am trying to better understand it, that is the reason for my post. I thought the condition for a unique lifting was $f_{\#}\pi_{1}(Z,x) \subset H(\tilde{x})$ basically $f$ sharp induces a group that is strictly a subgroup of $H$. However, that post contends that $S^{2}$ (a space that is simply connected there for has trivial fundamental group) lifts uniquely to $R^{2}$, but in this case, isn't $H$ the trivial group as well? So it would not be a proper subgroup.

Tim
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George Babus
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    It doesn't have to be a proper subgroup. Note that some people use $A \subset B$ to denote that $A$ is a (not necessarily proper) subset of $B$. – Michael Albanese Jul 27 '23 at 17:20

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