Let $f:T\to S$ be a morphism of base schemes and $q_X:X\to T$ a $T$-scheme. Assume the Weil restriction $R_{T/S}(X)$ exists, i.e. there is an $S$-scheme $R_{T/S}X\to S$ such that $$ \hom_T(U\times_S T,X) \simeq \hom_S(U,R_{T/S}(X)), $$ natural in $U$.
Is there always a map $u$ which fits in to the following diagram? $\require{AMScd}$ \begin{CD} X @>{u (?)}>> R_{T/S}(X)\\ @V{q_X}VV @VVV \\ T @>{f}>> S \end{CD} Where does it come from?
Notes:
- This question is somehow dual to this question.
- For a long time, I thought that I could just start from the pullback diagram \begin{CD} X\times_S T @>{p_1}>> X\\ @V{p_2}VV @V{q_X}VV \\ T @>{f}>> S \end{CD} and apply the defining property (i.e. the adjunction between base change and Weil restriction) to $p_1\in\hom(X\times_S T,X)\simeq \hom(X,R_{T/S}X)$. But it seems this doesn't work because the little subscript requires you to have a morphism of $T$-schemes to begin with and $p_1$ is not $T$-linear, in fact $X$ is not even a $T$-scheme.
- The question may be more category-theoretical in flavor then scheme-theoretic, in fact you can ask the same question about an arbitrary category a pullback functor $f^*:\mathcal C_{/S}\to\mathcal C_{/T}$ and a (perhaps partial) right adjoint $f^*\dashv f_*$. There is always a $\mathcal C$-arrow $f^*X\to X$, but is there also a $\mathcal C$-arrow $X\to f_*X$?
Let's compose $q_X$ with $f$, this makes $X$ an $S$-scheme. Taking $U$ to be $X$ with this $S$-scheme structure we have one of two canonical projections $p_1 \in \hom_T(X\times_S T,X)$, from the commutative diagram of this fibered product it is easy to see that this projection is indeed $T$-linear. Map $u \in \hom_S(X,R_{T/S}(X))$ is the map corresponding to this projection under given bijection, and your commutative diagram for $u$ just means that $u$ an $S$-linear map, where $X/S$ is as above.