The Proposition is that : Let $f:X\longrightarrow Y$ be a flat morphism of schemes of finite type over a field $k$. For any point $x\in X$, let $y=f(x)$. Then $\dim_x(X_y)=\dim_x(X)-\dim_y(Y)$. Here for any scheme $X$ and any point $x\in X$, by $\dim_x(X)$, we mean the dimension of the local ring $\mathcal{O}_{x,X}$.
They begin the proof as follows :
First we make a base change $Y'\longrightarrow Y$ where $Y'=\textrm{Spec } \mathcal{O}_{y,Y}$ and consider the morphism $f':X'\longrightarrow Y'$ where $X'=X\times_{Y} Y'$. Then $f'$ is also flat, $x$ lifts to $X'$ and the three numbers are the same.
What is meant by : $x$ lifts to $X'$. It is not the inverse image, because the inverse image could contain more than one element. What does it mean?
Thank you in advance!
@poorna The answer to your question follows from the universal property of the fibre product:
We have the fibre product
$$\begin{array} X\ \ \ X' & \stackrel{}{\longrightarrow} & X \\ \ \ \downarrow{f'} & & \downarrow{f} \\ Spec(\mathscr O_{Y,y}) & \stackrel{j_y}{\longrightarrow} & Y \end{array} $$
Consider the local ring $A = Spec(\mathscr O_{X,x})$. Due to the universal property of the fibre product the commutative square defining in a topological sense $x$ and $y=f(x)$
$$\begin{array} XSpec \ A & \stackrel{j_X}{\longrightarrow} & X \\ \ \ \downarrow{} & & \downarrow{f} \\ Spec(\mathscr O_{Y,y}) & \stackrel{j_y}{\longrightarrow} & Y \end{array} $$
maps into the fibre product via a unique morphism
$$Spec \ A \stackrel{j_{x'}}{\longrightarrow} X'$$
Referring to my previous comment, the latter morphism is a point of $X'$. It lifts the 'point' $x$, i.e. $Spec \ A \stackrel{j_X}{\longrightarrow} X$, to $X'$.