I need help in this definition of morphism of affine algebraic sets which I found in a book:
Let $X$ and $Y$ affine algebraic sets and say
$$f:X\to X'\ \text{and}\ g:Y\to Y'$$
isomorphisms with $X'$ and $Y'$ affine closed sets.
A morphism $h:X\to Y$ correspond to a morphism:
$h':X'\to X\to Y\to Y'$
Where the map above are respectively $f^{-1}$, $h$ and $g$.
I didn't understand this definition, the author define $h$ using $h$?
I'm a little confused.
Thanks in advance
There is a bijection between functions $h: X \rightarrow Y$ and functions $h': X'\rightarrow Y'$.
Given $h: X\rightarrow Y$, we can compose with the isomorphisms $f^{-1}$ and $g$ to get $g\circ h\circ f^{-1}: X'\rightarrow Y'$.
Conversely, given $h': X'\rightarrow Y'$, we can compose with the isomorphisms $f$ and $g^{-1}$ to get $g^{-1}\circ h'\circ f: X \rightarrow Y$.
$$\begin{array} XX & \stackrel{h}{\longrightarrow} & Y \\ \updownarrow{f} & & \updownarrow{g} \\ X' & \stackrel{h'}{\longrightarrow} & Y' \end{array} $$
Presumably you already know what a morphism of affine closed sets is (a function which is polynomial in coordinates). The author is declaring that a function $h': X' \rightarrow Y'$ is a called a morphism (of algebraic sets) if and only if the corresponding function $h: X\rightarrow Y$ is a morphism (of closed sets).