Problem 3.1.2 in Liu -- Omission in problem statement? Exercise 3.1.2 in Liu's Algebraic Geometry and Arithmetic Curves is as follows. 

Let $f:X\rightarrow Y$ be a morphism of schemes. For any scheme $T$,
  let $f(T):X(T)\rightarrow Y(T)$ denote the map defined by
  $f(T)(g)=f\circ g$. Shows that $f(T)$ is bijective for every $T$ if
  and only if $f$ is an isomorphism (use the identity morphisms on $X$
  and $Y$).

I feel this is poorly stated. In order to have sections over a scheme $T$, don't $X$ and $Y$ need to be $T$-schemes. That is, don't we need structure morphisms $X\rightarrow T$ and $Y\rightarrow T$, and for $f$ to be compatible with these? Or am I misunderstanding something? What should the correct wording be?
 A: I don't see any reason why you can't or shouldn't define $X(T)$ to be the set of morphisms $T\to X$.  If you really want your schemes to be over something, remember that all schemes are defined over Spec$(\mathbb Z)$.
A: Andreas Blass is correct. For arbitrary schemes $X$ and $T$, $X(T):=\mathrm{Hom}_{\mathrm{Sch}}(T,X)$, and since every scheme admits a unique morphism to $\mathrm{Spec}(\mathbf{Z})$, the category of schemes is the same as the category of $\mathbf{Z}$-schemes, and $\mathrm{Hom}_{\mathrm{Sch}}(T,X)=\mathrm{Hom}_{\mathrm{Sch}/\mathbf{Z}}(T,X)$. Liu's statement is true and correct as stated. If you want to consider an arbitrary base, the statement would be: a morphism $f:X\rightarrow Y$ of $S$-schemes ($S$ being the base) is an isomorphism if and only if the induced map $\mathrm{Hom}_{\mathrm{Sch}/S}(T,X)\rightarrow\mathrm{Hom}_{\mathrm{Sch}/S}(T,Y)$ is bijective for all $S$-schemes $T$. These $\mathrm{Hom}$ sets are also usually denoted $X(T)$ and $Y(T)$ if it is understood that you are working in the category of $S$-schemes. 
