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!
Best Answer
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.