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I'm trying to understand applications of the homotopy lifting theorem, stated as follows:

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I have seen an application that if two loops $l$ and $l′$ based at $b$ in $X$ are homotopic, then by the above theorem they can be lifted to homotopic paths in $\tilde{X}$.

I don't fully understand how this follows. I understand that $l$ and $l′$ both lift to paths $\tilde{l}$ and $\tilde{l'}$ respectively in $\tilde{X}$ (via another theorem about path lifting). Say that $H$ is the homotopy between $l$ and $l′$, then I can see that by the above theorem $H$ has a unique lift $\tilde{H}$ such that $\tilde{H}$ restricted to $Y$ $\times$ {0} = $\tilde{l}$. But for $\tilde{H}$ to be a homotopy between $\tilde{l}$ and $\tilde{l'}$ don't we need that $\tilde{H}$ restricted to $Y$ $\times$ {1} = $\tilde{l'}$ also? How does this follow from the theorem as well?

Thank you!

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Note that in any case $\tilde{H}(-,1)$ is $some$ path in $\tilde{X}$ so by the uniqueness of path lifting (which could be interpreted as this theorem with $Y$ being a point) we only need to show that $\tilde{H}$ is a homotopy of paths in $\tilde{X}$. In this case $\tilde{H}(-,1)$ will be the unique lift of $l'$ with starting point $\tilde{l}(0)$. For this we only need to show that $\tilde{H}(0,-)$ respectivly $\tilde{H}(1,-)$ are just fixed points. To this end choose a covering neighbourhood $\varphi : \tilde{l}(0) \in U \mapsto V \ni b$ and note that by definition of a lift $\varphi \circ \tilde{H}(0,-) = H(0,-) = b$. Since $\varphi$ is bijective this proofs that $\tilde{H}(0,-)$ is constant and analagously for the end point.

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  • $\begingroup$ Thanks for your reply! Why is it the case that if $\tilde{H}$ is a homotopy of paths in $\tilde{X}$ then $\tilde{H}$(-,1) will be the unique lift of $l'$? $\endgroup$
    – IanR_314
    Commented May 2, 2020 at 10:58
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    $\begingroup$ Sorry, I should have probably made that clearer: $\tilde{H}$ is a lift of $H$, so by the very definition $\varphi \circ \tilde{H} = H$. In particular $\varphi \circ \tilde{H}(-,1) = H(-,1) = l'$, i.e. $\tilde{H}(-,1)$ is a path lift of $l'$, which is unique once we determine a starting point in the fiber over $b$. But it is part of my proof that this starting point must be $\tilde{l}(0)$. $\endgroup$ Commented May 2, 2020 at 16:34
  • $\begingroup$ Brilliant, thank you; I understand the strategy now. A final question - do we need to make sure the covering neighbourhood $U$ that is chosen contains all the points in $\tilde{H}(0,-)$ in order for the line $\varphi$ $\tilde{H}(0,-)$ = $H(0,-)$ = $b$ to make sense? If so, how can we do this? Thank you! $\endgroup$
    – IanR_314
    Commented May 4, 2020 at 11:16
  • $\begingroup$ Nice question actually. A priori we do not know. However the preimage of $V$ under $\varphi$ decomposes into disconnected open sets that map homeomorphically onto $V$ ($U$ is one of them). The equation $\varphi \circ \tilde{H}(0,-) = H(0,-)$ only makes sense on $\varphi^{-1}(V)$, but we know that $H(0,-)$ is just a point and hence completly contained in $V$. Thus $\tilde{H}(0,-) \in \varphi^{-1}(b)$. But since these are discrete points in different disconnected components by assumption $\tilde{H}(0,-)$ is indeed constant. $\endgroup$ Commented May 4, 2020 at 16:15

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