Under what condition does $f(f^{-1}(f(A))) = f(A)$? Basic question regarding function.
Let $f: X \to Y$, then for what $f$ does $f(f^{-1}(f(A))) = f(A)$? hold?
Obviously this relationship holds when $f$ is a bijection. 
This does not hold when $f$ is pure surjection because the inverse does not exist.
Does this also hold when $f$ is an pure injection? I think so. 
 A: Actually it always holds.
But $f^{-1}$ must not be considered as a function $Y\to X$. Instead, $f^{-1}(B)$ is defined for every $B\subset Y$ as
$$ f^{-1}(B) = \{x\in X| f(x)\in B\}.$$
With this definition, for any $x\in A$, $x\in f^{-1}(f(A))$, so that the inclusion $f(A)\subset f(f^{-1}(f(A)))$ holds. Moreover, if $x\in f^{-1}(f(A))$, then $f(x)\in f(A)$, and thus the other inclusion holds. Thus we have the equality $f(A)= f(f^{-1}(f(A)))$.
A: When $f$ is an injection function then $f^{-1}(f(A))=A$, as a result
$$f(f^{-1}(f(A))) = f(A)$$
A: I would assume here that $f^{-1}(S)$, for $S \subseteq Y$, is referring to the preimage of $S$ under $f$. To avoid confusion with the inverse function $f^{-1} : Y \to X$ (should such a thing exist), I'll follow the somewhat common convention to use square brackets when taking preimages, writing $f^{-1}[S]$ for $\{x \in X : f(x) \in S\}$ (this helps prevent mistaking inverse functions for preimages).
$f^{-1}[f(A)]$ is by definition the set $\{x \in X: f(x) \in f(A)\}$. Applying $f$ to this set (that is, to each element in the set) yields $$f(f^{-1}[f(A)]) = \{f(x) : f(x) \in f(A)\},$$
and this is $f(A)$.
