Can a specific function with different continuities be used to compare topologies?

Question

I am studying general topology using Munkres, currently reading part 1, chapter 2, section 19(The Product Topology). I came across an example that gives a function which is continuous w.r.t. the product topology but failed to be continuous w.r.t. to the box topology, and we know that the product topology is generally coarser than the box topology(further more, strictly coarser on infinite sets). I feel like there is a connection between them(maybe there is not, of course). I wonder if a specific function with different continuities on different topologies can be used to compare topologies. So here is the generalized statement:
Suppose $X$ is a nonempty set with 2 topologies, namely $\mathcal{T_1}$ and $\mathcal{T_2}$. $\mathcal{T_1}$ is finer than $\mathcal{T_2}$(possibly equal). If there is a function $f:A \rightarrow X$ ($A$ is an arbitrary set with topology $\mathcal{T}$) which is continuous with respect to $\mathcal{T_2}$ but failed to be continuous with respect to $\mathcal{T_1}$, can we claim that $\mathcal{T_1}$ is strictly finer than $\mathcal{T_2}$?
Is there any thorem or counterexample? Any help is appreciated. Thank you in advance.

Answer 1

Yes, we can. Because the alternative is that $\mathcal T_1=\mathcal T_2$, in which case a function $f\colon(A,\mathcal T)\longrightarrow(X,\mathcal T_1)$ if and only if it is continuous as a function from $(A,\mathcal T)$ into $(X,\mathcal T_2)$.