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I believe that $f^{-1}(\Delta) \neq Y$. Let $Y$ be a torsor for an elliptic curve $E$ over $\mathbb Q$ and choose $k$ so that $Y_k \cong E_k$ or in other words $Y(k) \neq \emptyset$. Choose the torsor so that the diagram $C_k\cong E_k\to \mathbb P^1_k$ commutes. I believe this can be done by letting the torsor correspond to a cocycle in $H^1(G_\mathbb Q, E[2]) \subset H^1(G_\mathbb Q, E)$.

In general, for a $\mathbb Q$ scheme $S$, from the functor of points persepective, $$f^{-1}(\Delta)(S) = \{\lambda: S_k \to Y_k : S_k\to Y_k\to X_k \text{ is defined over } \mathbb Q\}.$$

In our particular case, let's take $S = \mathbb Q$. Then, there are no maps from $S \to Y$ but there are maps from $S \to f^{-1}(\Delta)$: Simply take the origin for instance.

From the functor of points, it is at least clear that $Y$ sits inside $f^{-1}(\Delta)$.

Welcome to mathoverflow, Pat!

I believe that $f^{-1}(\Delta) \neq Y$. Let $Y$ be a torsor for an elliptic curve $E$ over $\mathbb Q$ and choose $k$ so that $Y_k \cong E_k$ or in other words $Y(k) \neq \emptyset$. Choose the torsor so that the diagram $C_k\cong E_k\to \mathbb P^1_k$ commutes. I believe this can be done by letting the torsor correspond to a cocycle in $H^1(G_\mathbb Q, E[2]) \subset H^1(G_\mathbb Q, E(\overline{\mathbb Q}))$.

In general, for a $\mathbb Q$ scheme $S$, from the functor of points persepective, $$f^{-1}(\Delta)(S) = \{\lambda: S_k \to Y_k : S_k\to Y_k\to X_k \text{ is defined over } \mathbb Q\}.$$

In our particular case, let's take $S = \mathbb Q$. Then, there are no maps from $S \to Y$ but there are maps from $S \to f^{-1}(\Delta)$: Simply take the origin for instance.

From the functor of points, it is at least clear that $Y$ sits inside $f^{-1}(\Delta)$. I am not sure what else is there.