is $x=y^2$ a function? A function is a relation that associates each element of one domain to one and exactly one element in it's co-domain.
This is the definition of a function I know.
But if I take the function $x=y^2$ this condition is not satisfied.
For $x=1$ I get $y=1$ and $y=-1$.
Does this $x=y^2$ is not a function?
and is this same as $y=\sqrt x$ and is $y=\sqrt x$ not a function too?
 A: You are correct. If $x$ is the independent variable and $y$ is the dependent variable, then $x=y^2$ is not a function. It's a tilted parabola, and fails the vertical line test. And you give a good example. If $x=1$, then we have two $y$-values: $-1$ and $1$. So, one point on the $x$ axis maps two points to the $y$-axis. To address your question on $y=\sqrt{x}$, notice that $\sqrt{x}$ must be positive, by definition. In other words, it's only the top half of your sideways parabola. The equation $x=y^2$ is BOTH curves $y=\sqrt{x}$ and $y=-\sqrt{x}$ together. These two are individually functions, but when you plot them on the same coordinate plane, the curve they make ($x=y^2$) is not a function. So: $x = y^2$, not a function. $y=\sqrt{x}$ is a function, because we take the square root to always be positive. And $x=y^2$ is both curves $y=\sqrt{x}$ and $y=-\sqrt{x}$ plotted together.
A: When you write down $x = y^2$, without any further information, it is assumed that you are referring to the set of points $(x,y) \in \mathbb{R}^2$ that satisfy that relation. Knowing wether or not this defines $x$ as a function of $y$, $y$ as a function of $x$, and at which points, is a job for the implicit function theorem. In this case, all ambiguity vanishes when you specify which one is the dependent variable or, in other words, set the domain and the numeric rule for computing values of your function.
If you are considering $x$ as a function of $y$, the proposed relation defines a function. Keep in mind that calling the variables $x,y$ means nothing with regards to which are the vertical and horizontal axis...
