If a non-constant function is continuous wrt two different topologies, can anything be said about how these two topologies are related?

I apologize in advance for my gross mathematical ignorance.

Let $X$ and $Y$ be sets, let $fcolon Xto Y$ be a non-constant function, and let $T_y$ be a topology on $Y$ and $T_x, T’_x$ be two different topologies on $X$. Assume that $f$ is continuous both as a function from $(X,T_x)to(Y,T_y)$ and $(X,T_x’)to(Y,T_y)$. Is it possible to relate $T_x$ and $T_x’$ to each other in any way?

My hope is to apply such a result as follows: Let $(X,T_x),(Y,T_y)$ be two topological spaces and consider the problem of determining whether or not a difficult to explicitly analyze function $fcolon Xto Y$ is continuous wrt $T_x$. First, let $T_x’$ be the initial topology on $X$ induced by $f$. There are two scenarios that might occur, depending on the answer to the question:

1. Continuity is related to some topological property biconditionally (best case): $fcolon(X,T_x)to Y$ is continuous iff properties $u_1,u_2,dots,u_n$ hold on both $T_x$ and $T_x’$ (e.g, perhaps some structure of the fundamental group, an invariant, etc.). Then we can possibly determine the continuity of $f$ by answering whether or not $T_x$ and $T_x’$ share the properties in question.

I am just as happy if there is no biconditonal, but the conditional is of the form "$T_x$ and $T_x’$ share properties $u_1,dots,u_n$ implies $f$ is continuous on $T_x$ iff $f$ is continuous on $T_x’$.

2. Continuity is conditionally related to properties $u_1,dots,u_n$ in only one way: If $f$ is continuous on both $T_x$ and $T_x’$ then, $u_1,dots,u_n$ hold on $T_x$ if and only if they hold on $T_x’$.

This is still useful, because in the case where we can verify that one or more of the $u_i$ fails to hold on both spaces, then the function cannot be continuous on both, and hence is not continuous on $T_x$.

This may be either a totally trivial question or a subject of vast literature. I deeply apologize that I know very little topology and cannot answer that for myself.