Is square root of $y^2$ for every $y>0,y\in\mathbb{R}$?

Question

I mean I got really confused by this question, Let's assume that we are considering $y=-2$ so $y^2 = 4$ and so its square root is $2$, but we also have an exponential operation where, $(a^m)^n=a^{mn}$. So by using that when $y=-2, (y^2)^\frac12=y^{\left(2\cdot\frac12\right)}=y=-2$ so, basically $2=-2$? What is wrong here? Will that exponent operation not work in this operation?

Answer 1

What's wrong is that the equality$$a^{xy}=(a^x)^y$$doesn't hold without restrictions. It holds when $a>0$ and $x,y\in\mathbb R$, but not in general.