This might appear silly, I know that
$$ \begin{array}{l} f(A) = \left\{f(x) : x \in A \right\} \\ f^{-1}(A) = \left\{x : f(x) \in A \right\} \end{array} $$
If $y\in f(A)$ is fixed can this be expressed by the condition
$$ \exists x, x\in A \Rightarrow f(x)=y $$
Can we do a similar thing with the pre-image?
Basically I'd like to express those definition using logical connectors. Is this possible?
$y\in f(A)$ is equivalent to $$ \exists x\,(x\in A\land f(x)=y)$$ and $x\in f^{-1}(A)$ is equivalent to $$ f(x)\in A.$$
$y\in f(A)$ is equivalent with: $\exists x\;[x\in A\wedge f(x)=y]$.
So $\wedge$ instead of $\implies$.
We have the (very useful) equivalence:$$x\in f^{-1}(A)\iff f(x)\in A$$
So if $x\in f^{-1}(A)$ is fixed then this can be expressed by the condition $f(x)\in A$.
Does this meet your needs?
