Are the lifts of homotopic curves necessarily homotopic? Studying for an exam I fail to understand (this among many things) the connection between the homotopy of functions to that of their lifts.
Let $X, Y$ be path-connected topological spaces.
$p:X\to Y$ is a covering map, and $f,g:[0,1]\to Y$ are two loops with same origin such that $f(0)=f(1)=g(0)=g(1)=y_0$,
$\hat f,\hat g:[0,1]\to X$ their lifts with $\hat f(0)=\hat g(0) = x_0$
Is the fact that $[f]=[g]$ (homotopy relation) means $[\hat f]=[\hat g]$ (homotopy of lifts)?
What are the conditions for that to happen, and why?



*

*I struggled a bit trying to define a homotopy between the two lifts, but I needed some inverse projection from $Y$ to $X$. meaning - I can project $\hat f$ using $p$ to get $p\circ \hat f=f$, then deform it using the given homotopy $H$ until I reach $g$, but then what?

*I know the lift is unique (given an anchor $x_0$), so there is a function $h$ that send a curve to its lift anchored in $x_0$, but how can tell if that function is continuous (or having some other interesting property)? I such $h$ was continuous, I could maybe use it to define the mentioned homotopy...
 A: This always happens and is due mostly to the homotopy lifting property a property which covering spaces are guaranteed:

Given a covering space $p:\tilde X \rightarrow X$, a homotopy $f:Y \times I \rightarrow X$ and a map $\tilde f_0:X \rightarrow E$ lifting $f_0$ then there is a unique homotopy $\tilde f:Y \times I \rightarrow E$ of $\tilde f_0$ that lifts $f$.

The proof should be in your course text but goes something like this. For each point $y\in Y$ you build a lift $\tilde f: N\times I \rightarrow E$ for a small neighborhood $N$ of $y$ then notice each $f(N \times I)$ is contained in some $U_i$ from the corresponding open cover $\{U_{\alpha}\}$ of $X$ that gets disjoint homeomorphic copies by $p^{-1}(U_i)$. So that $\tilde f(N\times I)$ must be in one of these sheets and hence for overlapping $N$, $\tilde f$ agrees and we build a lift on all $Y\times I$.
To finish by directly addressing your situation, you have $[f] = [g]$ so we have a homotopy $F:I\times I \rightarrow Y$ with $F(t,0) = f$, $F(t,1)=g$ and further a lift $\hat f$ of $f$ then by the homotopy lifting property we have a homotopy $\tilde F:I\times I \rightarrow X$ that lifts $F$. Giving us $\tilde g = \tilde F(t,1)$ is a lift of $g$ with $[\hat f] = [\tilde g]$. But how do we know $\tilde g = \hat g$ the given lift of $g$? Because of a second easily proven theorem about lifts: 

Given a covering space $p:\tilde X \rightarrow X$ and map $f:Y \rightarrow X$ with two lifts $\tilde f_1,\tilde f_2:Y \rightarrow \tilde X$ that agree at one point, then if $Y$ is connected they must agree on all $Y$.

Now with the fact that $I$ is connected and since $\hat g(0) = x_0 = \hat f(0) = \tilde F(0,s) = \tilde g(0)$ we have that $\hat g = \tilde g$ and finally $[\hat f] = [\hat g]$.
