# Understanding homotopy lifting

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

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!

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.
• 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 $l'$? Commented May 2, 2020 at 10:58
• 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 $\tilde{l}(0)$. Commented May 2, 2020 at 16:34
• 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! Commented May 4, 2020 at 11:16
• 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. Commented May 4, 2020 at 16:15