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.