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The homotopy lifting property says that in case $a:I\rightarrow Y$ is homotopic to $b:I\rightarrow Y$ (denoting the homotopy in $F:I\times[0,1]\rightarrow Y )$ and $p:X\rightarrow Y$ is a covering map, for $\tilde{a} : I\rightarrow X$ a lifting of $a$ ($(p\circ\tilde{a}) (s)= a(s)$) then there exists a unique homotopy $G:I\times[0,1]\rightarrow X$ such that $p \circ G(s,t) = F(s,t)$. So as I understand it, it means that for $\tilde{a}$ there exists a unique lifting $\tilde{b}$ of $b$ such that $\tilde{a} \simeq \tilde{b}$.

I try to find an example of $a\simeq b$ such that $\tilde{a}$ is not homotopic to $\tilde{b}$, My intuition was creating a covering map: $p:X\rightarrow Y$ such that two homotopic loops $a,b$ in $(I,\partial I)\rightarrow (Y,y)$ where $a$ is lifted to a loop $\tilde a :(I,\partial I)\rightarrow (X,x\in P^{-1}(y))$ and $\tilde b :I\rightarrow X$ is not a loop in $X$ ($\tilde{b(0)},\tilde{b(1)}\in P^{-1}(y)$). Yet I faild to construct such spaces and covering map.

Any idea would be appreciated.

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    $\begingroup$ If I understand correctly what you are looking for then you may consider the trivial covering, with $X$ of the form, say, $X\simeq Y\times\mathbb{Z}_2$. Then lift $a$ into $Y\times 0\subseteq Y\times\mathbb{Z}_2$ and lift $b$ into $Y\times 1$. $\endgroup$ – Tyrone Aug 19 '18 at 10:02
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A covering projection $p : X \to Y$ is a fibration which means that it has the homotopy lifting property. This does neither mean that any map $f : Z \to Y$ has a lift $\tilde{f} : Z \to X$ nor that lifts are unique. We can only say that if $Y$ is connected and two lifts agree for some point of $Z$, then they are identical.

In particalar you cannot say that the homotopy $F$ has a unique lift $G$, but only that there exists a unique lift $G$ such that $G_0 = \tilde{a}$. In other words, you get a lift $\tilde{b}$ of $b$ such that $\tilde{b} \simeq \tilde{a}$, but in general there are other lifts which are not homotopic to $\tilde{a}$. See Tyrone's comment.

By the way, you have to make precise what you mean by a homotopy of paths. The usual approach is to consider only paths $u, v : I \to Y$ such that $u(0) = v(0) = y_0$ and $u(1) = u(1) = y_1$ and homotopies keeping $y_0, y_1$ fixed. The set of such homotopy classes is called the fundamental grupoid of $Y$.

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  • $\begingroup$ I am sorry for down voting, but for your second claim - I said that for a certain lift of $F(s,0)$ there is a unique G which satisfies $p\circ G(s,t) = F(s,t)$ .Which is as much as I'm concerned the same thing as you claimed. For the third part, your description is for relative homotopy means $v \simeq _{rel \partial{I}} u$ but all I meant was the basic definition for homotopy, relating $a,b$ (paths) as functions: $I \rightarrow Y$. $\endgroup$ – dan Aug 20 '18 at 4:27
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    $\begingroup$ @Dan Downvoting is okay if you are not satisfied with an answer, However, in your question you did not mention that you require $G(s,0) = \tilde{a}$ (probably this was obvious to you). $\endgroup$ – Paul Frost Aug 20 '18 at 12:56

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