The diagonal of the Weil restriction

As with pretty much everything to do with Weil restrictions, it becomes more clear if we base change everything to an algebraically closed field $\overline{\mathbb Q}$. The base change of $k$ is just $\overline{\mathbb Q}^n$ for $n$ the degree of $k$ over $\mathbb Q$. Then the base change of the Weil restriction from $k$ to $\mathbb Q$ $Y$ is the Weil restriction of $Y$ from $\overline{\mathbb Q}^n$ to $\overline{\mathbb Q}$, in other words, $Y^n$. Similarly the base change of the Weil restriction of $X$ is $X^n$. So the inverse image is the inverse image of the diagonal $X$ inside $Y^n$, which is $Y \times_X Y \dots \times_X Y$.

So the geometric irreducible components will be the geometric irreducible components of that fiber product, and the arithmetic irreducible components will be the orbits on the geometric irreducible component of $\operatorname{Gal}(\overline{\mathbb Q} / \mathbb Q)$, which is the obvious action twisted by the action permuting the $n$ embeddings of $k$ into $\overline{\mathbb Q}$.

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.