Let $X,Y$ be sets, and let $\pi_{X×Y \to X} : X×Y \to X$ and $\pi_{X×Y→Y} : X × Y \to Y$ be the maps $\pi_{X×Y→X}(x,y) := x$ and $\pi_{X×Y}→Y (x,y) :=y$; these maps are known as the co-ordinate functions on $X × Y$.
Show that for any functions $f : Z \to X$ and $g : Z \to Y$ , there exists a unique function $h : Z → X×Y$ such that $\pi_{X×Y→X}◦h = f$ and $\pi_{X×Y→Y}◦h = g$.
I don't know how to solve the problem. Below are just some thoughts of mine:
$z \in$ $\pi_{X×Y→X}◦h = f$ and $\pi_{X×Y→Y}◦h = g$ if
$z \in$ $\pi_{X×Y→X}◦h(z) = f(z)$ and $\pi_{X×Y→Y}◦h(z) = g(z)$
$z \in$ $\pi_{X×Y→X}◦h(z) = f(z) = x \in X$ and $\pi_{X×Y→Y}◦h(z) = g(z) = y \in Y$
$\pi_{X×Y→X}◦h = f$ and $\pi_{X×Y→Y}◦h = g$ $\subset X \times Y $ Don't see any way to proceed...
Suppose there is no $h : Z → X×Y$ such that $\pi_{X×Y→X}◦h = f$ and $\pi_{X×Y→Y}◦h = g$.
That is for any $z \in Z$ we get $\pi_{X×Y→X}◦h(z) = \pi_{X×Y→X}(x_1,y) = x_1; f(z) = x_2; x_1 \not=x_2$ or $\pi_{X×Y→Y}◦h(z) = \pi_{X×Y→Y}(x, y_1) = y_1; g(z) = y_2; y_1\not=y_2$. Don't see any way to continue...
The conditions with the coordinate functions (which I'll call $\pi_X$ and $\pi_Y,$ instead, for brevity) can be translated as follows:
$\pi_X\circ h=f$ means that for any $z\in Z,$ if $h(z)=(x,y),$ then $x=f(z).$
$\pi_Y\circ h=g$ means that for any $z\in Z,$ if $h(z)=(x,y),$ then $y=g(z).$
Thus, for any $z\in Z,$ if $h(z)=(x,y),$ then $h(z)=(?,?).$ Can you take it from there to see how $h$ must be defined?