This question is related to this one but I have written this in a self-contained manner.

All varieties are complex varieties.

For quasi-projective variety $U$ and a projective variety $X$ we can topologize the space of regular maps from $U$ to $X$ using the topology of convergence by the bounded degree. For this purpose we need a projective closure $\overline{U}$ of $U$. This topology is independent of choice of the closure. A sequence $f_i$ is convergent iff it converges in $\text{Hom}_{\text{cont}}(U^{an}, X^{an})$ with the compact open topology and there is an upper bound on the degree of the closure of the graphs of $f_i$'s in $\overline{U}\times X$. In the case $U$ is already projective this coincides with the topology on the complex points of the $\text{Hom}$ scheme. We denote this topological space by $\text{Mor}(U,Y)$

Now let's assume a finite group $G$ is acting on $X$ and the quotient denoted by $X/G$ is a projective variety. We have the following map induced by quotient of the group action:

$$f_G: \text{Mor}(U,X)\rightarrow \text{Mor}(U, X/G)$$

Let's denote the subspace of $\text{Mor}(U, X/G)$ consisting of liftable to $\text{Mor}(U,X)$ by $\text{Mor}_{\text{liftable}}(U, X/G)$.

Let's $\overline{K(U)}$ be the algebraic closure of the function field of $U$. We topologize the space of morphisms from $\overline{K(U)}$ to $X/G$ by taking the filtered colimit topology induced by all $Mor(U_{\alpha}, X/G)$ where $U_{\alpha}$ is the family of quasi-projective varieties consisting of connected etale opens of $U$ and the transition maps is given by restriction along these etale opens. We simply denote this topological space by $\text{Mor}(\overline{K(U)}, X/G)$.

Let's denote the topological space constructed by taking the filtered colimit of $\text{Mor}_{\text{liftable}}(U_{\alpha}, X/G)$ by $\text{Mor}_{\text{liftable}}(\overline{K(U)},X/G)$. There is a natural map $i: \text{Mor}_{\text{liftable}}(\overline{K(U)},X/G)\rightarrow \text{Mor}(\overline{K(U)}, X/G)$ which is a bijection on the point level. This is because any map from $\overline{K(U)}$ to $X/G$ can be lifted to $X$. Is $i$ necessarily a homeomorphism?