Show that continuity-preserving maps are homeomorphisms

Question

Let $X$ and $Y$ be topological spaces. Call $f:X \rightarrow Y$ continuity-preserving if, for all continuous functions $\gamma: \mathbb{R} \rightarrow X$, $f \circ \gamma$ is also continuous. I know that all continuous functions are continuity-preserving; that is, if $f$ is continuous and $\gamma$ is continuous then $f \circ \gamma$ is also continuous. But does the converse hold? That is, are all continuity-preserving curves continuous? And if so, is there a simple way of proving it?

Answer 1

No, it is not true. If $X$ is totally (path) disconnected, then only constant maps $\mathbb{R}\to X$ are continuous. And this implies that every function $X\to Y$ for any topological space $Y$ is continuity-preserving.

But it is not hard to find an example of a discontinuous map $X\to Y$ with $X$ totally disconnected, e.g. $X=\mathbb{Q}$.

This is true if $X$ is locally path connected first countable. In such scenario if $f:X\to Y$ is discontinuous then there is a sequence $x_n\to x$ such that $f(x_n)\not\to f(x)$. By the fact that $X$ is locally path connected, we can create a continuous path $\lambda:[0,1]\to X$ such that $\lambda(1)=x$ and (eventually) $\lambda(1-1/n)=x_n$. But then $f\circ\lambda$ remains discontinuous.