Continuity condition for homogeneous topological spaces I know that given topological groups $G_1,G_2$ and a homomorphism $f:G_1\to G_2$, $f$ is continuous if it is continuous at a point. If I am not mistaken, this follows using the fact that translation maps of the form $h\mapsto g\cdot h$, 'commute' with the map $f$ and are homeomorphisms. In the sense that
$\require{AMScd}$
$$ \begin{CD}
    \mathbb (x,y) @>\displaystyle{\cdot} >> x\cdot y\\
    @V \displaystyle f\times f V V\ @VV \displaystyle{f} V\\
    \mathbb (f(x),f(y) ) @>>\displaystyle \cdot> f(x\cdot y)
    \end{CD}$$
commutes.
My question is, are there natural generalizations to such results only using this property? i.e., given two topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$ with transitively acting groups $H_X<Homeo(X)$ and $H_Y<Homeo(Y)$, can we say that $f:X\to Y$, which is continuous at a point $x_0\in X$, is a continuous map in general if we have some approriate commutativity of $f$ with the actions?
I think that if one can show somehow that
$$ \begin{CD}
     x @>\displaystyle{f} >> f(x)\\
    @V \displaystyle \phi_x V V\ @AA \displaystyle{\psi_{f(x)}} A\\
    x_0 @>>\displaystyle f> f(x_0)
    \end{CD} $$
commutes in some sense, where $\phi_x\in H_X$ and $\psi_{f(x)}\in H_Y$ satisfy $\phi_x(x)=x_0$ and $\psi_{f(x_0)}\big( f(x_0) \big)=f(x)$, then it will solve the problem. Is there some reasonable conditions on group actions satisfies this?
 A: Continuity at points is preserved under composition:

*

*Given topological spaces $X, Y, Z$ with $x \in X$ and given maps $f : X \to Y, g: Y \to Z$, if $f$ is continuous at $X$ and $g$ is continuous at $f(x)$, then $g\circ f$ is continuous at $x$.

For, letting $y = f(x)$ and $z = g(y)$, if $U$ is a neighborhood of $z$, then $g^{-1}(U)$ is a neighborhood of $y$ by the continuity of $g$ at $y$, and by the continuity of $f$ at $x, f^{-1}(g^{-1}(U)) = (g\circ f)^{-1}(U)$ is a neighborhood of $x$. Since this holds for all neighborhoods of $z=(g\circ f)(x)$, $g \circ f$ is continuous at $x$.
Now for a topological space $X$, define an homogenization $\mathscr H_X$ (all related terminology is my own invention) of $X$ to be a collections of maps $\phi : U \in \tau_X \to X$ such that for every $x, y \in X$, there is a $\phi \in \mathscr H_X$ with $x \in \text{dom}(\phi)$, such that $\phi(x) = y$ and $\phi$ is continuous at $x$. Such a $\phi$ is said to represent $x$ to $y$.
For homogenizations $\mathscr H_X, \mathscr H_Y$ of $X, Y$, call $f:X \to Y$ compatible  with $\mathscr H_X, \mathscr H_Y$ if for all $x, z \in X$, there exist representations $\phi$ of $x$ to $z$ and $\psi$ of $f(z)$ to $f(x)$ such that
$$f = \psi \circ f \circ \phi$$
on some neighborhood of $x$.
Finally, we get the theorem:

*

*Given spaces $X, Y$ with homogenizations $\mathscr H_X, \mathscr H_Y$, if $f : X \to Y$ is compatible with $\mathscr H_X, \mathscr H_Y$, and if $f$ is continuous at some point $z \in X$, then $f$ is continuous on all of $X$.

If $x$ is any point of $X$, then there is a representation $\phi$ of $x$ to $z$, and a representation $\psi$ of $f(z)$ to $f(x)$ such that $f = \psi\circ f\circ\phi$ on some neighborhood of $x$. $\phi$ and $\psi$ are continuous at $x$ and $f(z)$ with respect to their domains, but since those domains are open sets in $X$ and $Y$ respectively, they are also continuous at these points with respect to the topologies of $X$ and $Y$. Hence by the earlier theorem and the continuity of $f$ at $z$, $\psi \circ f \circ \phi$ is continuous at $x$. Since it agrees with $f$ on some neighborhood of $x$, $f$ is continuous at $x$. As $x$ was arbitrary, $f$ is continuous at every point of $X$.

That is the most abstract form of the process that I can think of. Note that I asked far less of $\mathscr H_X$ than you. There is a reason for that. The fewer requirements I can place on $\mathscr H_X$, the bigger it can be, which allows more maps $f$ to be compatible with it, which makes the theorem more widely applicable.
If you want a more restrictive set of maps such as you have described, you just have to show that they none-the-less still satisfy the conditions to be an homogenization.
