What image of a (inverse) set implies Confused about this stuff. Just trying to organize it in my head.
Let $f: X \to Y$ with $A \subseteq X.$ Then $f(A) = \{f(x): x \in A\}$. 
Does this definition imply that $x \in A \to f(x) \in f(A)$ where the converse is only true if $f$ is injective? Or is it separate from the definition of $f(A)$?
Let $f: X \to Y$ with $C \subseteq Y.$ Then $f^{-1}(C) = \{x: f(x) \in C\}$. 
Does this definition imply $x \in f^{-1}(C) \iff f(x) \in C$ ? If so, is this "if and only if" statement just a definition or an actual theorem that needs to be proved?
 A: What is confusing you is your use of the implication. All of your statements are true, though.
If I define $A$ as $\{x \in \mathbb{N} | x^2 =64 \}$, then it is perfectly true to say $x \in A$ implies $x^2=64$ (your first question), and is due to the definition alone. 
It is also perfectly true to say $x^2=64$ if and only if $x \in A$ (your second question) and is also due to the definition alone
A: The set builder notation definitions of the two tell you the following:
$$x \in A \Rightarrow f(x) \in f(A) \\
x \in f^{-1}(B) \Leftrightarrow f(x) \in B.$$
Notably, $f^{-1}(B)$ is the image of $f^{-1}$ if $f$ is injective, but it is well-defined regardless.
The first implication is not an equivalence basically because the definition of $f(A)$ is not of the form $\{ x : P(x) \}$ for some predicate $P$, whereas this is true of $f^{-1}(B)$. Rewriting it in that form reveals the difference: $f(A) = \{ y \in Y : (\exists x \in X) \: f(x)=y \}$. This existential quantifier changes matters, because you could have $f(x) \in f(A)$ if $z \in A$ and $f(z)=f(x)$ even if $z \neq x$. Of course this cannot happen if $f$ is injective.
