Conditions equivalent to the surjectivity of a function 
Show that the following are equivalent for a map $f:X \to Y$.
  
  
*
  
*$\,f$ is a surjection.
  
*$\,f[f^{-1}[B]]=B$ for each $B \subseteq Y$.
  
*$\,f^{-1}[B] \subsetneq f^{-1}[C]$ for each $B \subsetneq C \subseteq Y$.
  

 A: Proof.
$1.\,\Longrightarrow\, 2.$ The "$\subset$" part holds always, even if $f$ is not surjective, since if $y\in f\big[f^{-1}[B]\big]$, then $y=f(x)$, for some $x\in f^{-1}[B]$, and thus $y=f(x)\in B$. For the "$\supset$" part, let $y\in B$.
Since $f$ is surjective, there exists an $x\in f^{-1}[B]$, such that $f(x)=y$, and thus, $y=f(x)\in f[f^{-1}[B]]$. 
$2.\,\Longrightarrow\, 3.$ Assume now that $B \subsetneq C \subset Y$. As it is clear that $f^{-1}[B]\subset f^{-1}[C]$, we need to show the strict inequality. Let $y\in C\smallsetminus B$. Since $f$ is surjective, there exists an $x\in f^{-1}[C\smallsetminus B]\subset f^{-1}[C\smallsetminus B]$, for which $f(x)=y$. But $x\not\in f^{-1}[B]$, because if $x\in f^{-1}[B]$, then $y=f(x)\in f^{-1}[B]$. Therefore $f^{-1}[B]\subsetneq f^{-1}[C]$.
$3.\,\Longrightarrow\, 1.$ Let $y\in Y$, we need to show, assuming 3.,  that there exist a $x\in X$, such that $f(x)=y$. Clearly $\varnothing\subsetneq \{y\}\subset Y$. Due to 3. $$
f^{-1}[\varnothing]\subsetneq f^{-1}[\{y\}]. \tag{1}
$$ 
But, as $f^{-1}[\varnothing]=\varnothing$, $(1)$ implies that  $f^{-1}[\{y\}]\ne\varnothing$. For every $x\in f^{-1}[\{y\}]$, we know that $f(x)=y$. And this concludes the proof.
A: I think, as a commenter pointed out, that your confusion stems from misunderstanding what $f^{-1}[B]$ means when $B \subseteq Y$.
Here, since we really don't know that the function $f$ is invertible, we can't assume that $f^{-1}$ stands for a function $ f^{-1} : X \leftarrow Y$. Instead, it means the preimage under $f$ of $B$. That is, the set of all $x \in X$ with $f(x) \in B$.
I think with this definition, you should be fine. Also, when you have to show the equivalence of $n > 2$ conditions, it's sometimes easier to try to prove $(1) \implies (2) \implies (3) \implies (1)$ (or some permutation thereof!) to save yourself from work.
