Lifting a map from the quotient by a contractible topological group

Question

Let $X$ be a topological space, $G$ be a contractible topological group acting freely and continuously on $X$, and $Y$ be the quotient. Let $A$ be a finite-dimensional CW-complex, and $f: A \rightarrow Y$ be a continuous map. Does this map always lift to $X$?

Answer 1

Here is a counterexample: any irrational flow on $X=S^1 \times S^1$. For example, start with the $\mathbb R$-action $r \cdot (x,y) = (x+r,y+\sqrt{2}r)$ on $\mathbb R^2$. Then mod out by the action of $\mathbb Z^2$ to get $S^1 \times S^1$ and an induced action of $\mathbb R$ on $S^1 \times S^1$ with quotient map $q : S^1 \times S^1 \to Y$. Every orbit of this action is a dense subset of $S^1 \times S^1$, and therefore the only saturated open subsets are the empty set and the whole of $S^1 \times S^1$. It follows that the quotient topology on $Y$ is the trivial topology. Every function $f : A \to Y$ is therefore continuous.

Now you can basically take $A$ to be any connected CW complex of positive dimension, and break it into two nonempty pieces. To be concrete I'll take $A=[0,1]$ to be the unit interval, broken into two pieces $[0,1]=[0,1/2] \sqcup (1/2,1]$.

Choose $x \ne y \in Y$, and define $f[0,1/2]=x$ and $f(1/2,1]=y$.

For any lift $\tilde f : Y \to S^1 \times S^1$ we have $\tilde f [0,1/2] \subset q^{-1}(x)$ which is one orbit of the flow, and $\tilde f(1/2,1] \subset q^{-1}(y)$ which is a different orbit of the flow.

But no such function $\tilde f$ can be continuous, for suppose it were. It would follow that for each $\delta>0$ the set $f(1/2,1/2+\delta)$ is a connected subset of the leaf $q^{-1}(y)$. Also, for each $\epsilon>0$ there would be a $\delta>0$ such that $f(1/2,1/2+\delta)$ is contained in the $\epsilon$-ball around $f(1/2)$ in $S^1 \times S^1$; in particular it follows that for each $\delta>0$ the set $f(1/2,1/2+\delta)$ would contain a point arbitrarily close to $f(1/2)$. However, if $\epsilon$ is sufficiently small then each connected component of $B \cap q^{-1}(y)$ has positive distance from $f(1/2) \in B \cap q^{-1}(x)$, which is a contradiction.

Answer 2

No, the lifting criterion gives the conditions necessary. That is that $f_*(\pi_1(A))\subset p_*(\pi_1(X))$. Where $p$ is the covering projection.

As a counter-example, take $X=S^2$ covering $Y=\mathbb{R}P^2$ under the action of the antipodal map, and then if $f$ simply assigns a CW-complex structure to $Y$, we have $\mathbb{Z}_2=f_*(\pi_1(A))\subset p_*(\pi_1(X))={id}$ which is your contradiction.