A less abstract description of Weil restriction $\DeclareMathOperator{\Hom}{Hom}$
$\DeclareMathOperator{\Res}{Res}$
See also this question.  Let $f: T \rightarrow S$ be a morphism of schemes.  Let $X$ be a $T$-scheme, and suppose the Weil restriction $\Res_{T/S}(X)$ exists.  That is, there is a an $S$-scheme $\Res_{T/S}(X)$ together with a collection of bijections
$$\tau_U:  \Hom_S(U,\Res_{T/S}(X)) \rightarrow \Hom_T(U \times_S T, X)$$
natural in $U$.  How can I get a more hands on description of this bijection, e.g. in terms of a morphism of $S$-schemes $X \rightarrow \Res_{T/S}(X)$?  I'm asking and answering this question as a future reference to myself.
 A: $\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Res}{Res}$It won't quite be that.  Let $\Phi: \Res_{T/S} X \times_S T \rightarrow X$ be the image of the identity map $1_{\Res_{T/S} X}$ under $\tau_{\Res_{T/S} X}$.  Let $U$ be any $S$-scheme.  
Apply the functors $\Hom_S(-,\Res_{T/S}X)$ and $\Hom_T(-\times_S T,X)$ to a given morphism $h: U \rightarrow \Res_{T/S}X$.  We obtain by base change a morphism of $T$-schemes $h \times 1_T: U \times_S T \rightarrow \Res_{T/S}X \times_ST$.On account of the commutativity of the (horrible) diagram
$$\begin{array} $\textrm{Hom}_S(\Res_{T/S} X, \Res_{T/S}X) & \xrightarrow{\tau_{\Res_{T/S}X}}  & \Hom_T(\Res_{T/S} X \times_S T,X) \\ \downarrow & & \downarrow \\
\Hom_S(U,\Res_{T/S}X) & \xrightarrow{\tau_U} & \Hom_T(U\times_ST,X)\end{array}$$
we begin with the identity map in the upper left corner, and obtain
$$\tau_U(h) = \Phi \circ (h \times 1_T)$$
Thus we have a morphism of $T$-schemes $\Phi:\Res_{T/S} X \times_S T \rightarrow X$ for which the map
$$h \mapsto \Phi \circ (h \times 1_T): \Hom_S(U,\Res_{T/S}X) \rightarrow \Hom_T(U \times_S T,X)$$
is bijective.
