Is the formula $\exists x [f(x) = a]$ logically valid?

Question

Since the variable $x$ belongs to the universal set, then we could say that $f^{-1}(a)$ is one of the values we could assign to $x$. Therefore, $\exists x [(f(x) = a]$ is equivalent to $f(x_0)=a \lor f(x_1)=a \lor \ldots f[f^{-1}(a)]=a \ldots f(x_n)=a$. Since $f[f^{-1}(a)]=a$ is true, then $\exists x [(f(x) = a]$ is always valid. Is it correct?

Answer 1

Of course not and you know it.

Let the universal set be $\mathbb R$ and $f(x) = x^2$ and $a = -1$. So is there an $x \in \mathbb R$ so that $x^2 = -1$? No.

Your error is when you say $f(f^{-1}(a)) = a$ always. That just is not true.

$f(f^{-1}(-1)) = (f^{-1}(a))^2 = \sqrt {-1}^2 \ne -1$ because $\sqrt{-1}$ does not exist so you can not square it.

More generally $f^{-1}(a) = \{x| f(x) = a\}$ could very well be empty.

And if so.... $f(f^{-1}(a)) = \{f(x)| x \in f^{-1}(a)\}= \{f(x)|x\in \emptyset\} =\emptyset$.

......

Okay. Let $U$ be your "universal set of unrestricted domain of discourse".

Now let $f(x) = \begin{cases}1& \text{if }x\in U\\x&\text{if }x \not \in U\end{cases}$

Then what is the solution to $f(x)= 2$?

As $f(x) \ne 1$ we must have $x \not \in U$. So that means $f(x) =x$ So $f(x) = x = 2$. So $2 \not \in U$. But $2\in U$ because surely a universal set of unrestricted domain can not be restricted to exclude $2$. A contradiction.

....

Or more bluntly. Let $f(x) = 1$. Let $x\in f^{-1}(2)$. Then $f(x) = 2$. But $f(x) =1$. A contradiction. Doesn't matter which domain we choose.