My understanding is that Reduce gives all conditions (using or) where the input is true.

Now, $ \sqrt{xy} = \sqrt x \sqrt y $ , where $ x,y$ are real, under the following three conditions/cases

$ $
\begin{align*}
x\geq 0,y\geq0\
x\geq0,y\leq0\
x\leq0,y\geq 0 \
\end{align*}
$ $

but not when $ x<0,y<0$

This is verified by doing

ClearAll[x,y]
Assuming[Element[{x,y},Reals]&&x>= 0&&y<= 0,Simplify[ Sqrt[x*y] - Sqrt[x]*Sqrt[y]]]
Assuming[Element[{x,y},Reals]&&x<= 0&&y>= 0,Simplify[ Sqrt[x*y] - Sqrt[x]*Sqrt[y]]]
Assuming[Element[{x,y},Reals]&&x<= 0&&y>= 0,Simplify[ Sqrt[x*y] - Sqrt[x]*Sqrt[y]]]
Assuming[Element[{x,y},Reals]&&x<= 0&&y<=  0,Simplify[ Sqrt[x*y] - Sqrt[x]*Sqrt[y]]]

Then why does

 Reduce[ Sqrt[x*y] - Sqrt[x]*Sqrt[y]==0,{x,y},Reals]

Give only one of the 3 cases above?

Is my understanding of Reduce wrong or should Reduce have given the other two cases?

V 12 on windows.

As Coolwater says in his comment, using the domain specification Reals means that all function values are constrained to be real. Clearly Sqrt[x] is not real when $ x<0$ . Instead, constrain x and y to be real using Element:

Reduce[Sqrt[x y] - Sqrt[x] Sqrt[y] == 0 && (x|y) ∈ Reals, {x,y}] //Simplify

(y ∈ Reals && x > 0) || x == 0 || (x <= 0 && y >= 0)