Perhaps this is a silly question, but if you have a function, such as
$$f(x) = \frac{x^2}{x}$$
the domain is all real numbers except x = 0.
However, this function simplifies to
$$f(x) = x$$
which has a domain of all real numbers. The domains for the two functions are different. Why are you permitted to simplify the first function if the domain changes?
To be strictly correct, the domain does not change. The simplified version of $$f(x)=\frac{x^2}{x}\ ,\quad x\ne0$$ is $$f(x)=x\ ,\quad x\ne0\ .$$
Really, a function is not properly specified unless the domain is stated. If a function is given without a domain then the normal default assumption is to take the largest possible domain. So, if you were given a proposed function $f(x)=x$ with no "back story" you would probably (but need not) choose the domain $\Bbb R$. But this case is different since you do have some prior information about the function.
Think of domain of more as a graph. If you have $y=x$ (as in your example), is the function defined at $0$? yes. If you have $y=\frac{x^2}{x}$ and you plug in $0$, it's an improper fraction, so it wouldn't even "work" unless it was simplified.
If you have something like $\frac{(x-3)}{(x^2-9)}$, $x$ can't be $3$, right? But this is a "legit" function at $x=3$ (undefined, rather than improper). So when you simplify something like this to $y=\frac{1}{(x+3)}$ the domain is STILL $x \neq 3$
