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I am in the following setting:

Let $X=[0,1]$, $Y=[0,1]^2$ and $Z=Y^n/\sim$, where $\sim$ is the equivalence relation given by $(x_1,\ldots,x_n)\sim (y_1,\ldots,y_n)$ if and only if there is a permutation $\sigma$ on $\{1,\ldots,n\}$ with $x_i=y_{\sigma(i)}$. Let $\pi\colon Y\to Z$ be the quotient map.

Does every path $X\to Z$ lift to a path $X\to Y$ making the diagram commute (with $\pi$)?

The problem is that $Y$ is not a covering space for $Z$, nor a fiber bundle (as the inverse image of $(x,\ldots,x)$ is only one point, but if all the $x_i$'s are different points in the square, then the inverse image of $(x_1,\ldots,x_n)$ consists of $n!$ many points) and all the lifting theorems I know do not apply.

Note, that this works if $Y$ is replaced with the interval itself (essentially because the topology on $[0,1]$ is an order topology). I was tempted to identify $Y^n$ with $[0,1]^{2n}$, but $\sim$ defines a weaker equivalence relation than the one defined by permutations on $[0,1]^{2n}$.

I've been trying to lifts paths in this particular case since a while ago, but unsuccessfully..

Thanks!

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