$\forall C\subset A, \; \forall D\subset B, \; f(C)\subset D \iff C\subset f^{-1}(D)?$ I've been asked to demonstrate this in my elements of mathematics class:

$$\forall C\subset A, \; \forall D\subset B, \; f(C)\subset D \iff C\subset f^{-1}(D)$$

I've made two trials:

  
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*First I've tried to use the definition of the inverse image, and then I wrote:
  
  
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*Considering the inverse image, given $D\subset B$.
  
  
  
  $$f^{-1}(D):=\{x\in A: f(x)\in D\}$$
With this I think that I've found the way back from the elements of $D$, but I can't isolate them in a subset $B$.

$ $


  
*Then I've tried to characterize the functions to find the subsets $C$ and $D$:
  
  
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*$f(x \in A)=y\in B$
  
*$f^{-1}(y\in B)=\{x:x\in A:f(x)\in D\subset B \}$
  
  
  
  With this I believe I've been able to characterize the set $B$, but somehow I can't do the reverse process to characterize the set $A$.

Also, the professor marked the exercise as important. But I have little clue of why it is important. 
Edit: I've tried something else:

In my professor notes, he gave the definition of function using relations, for $f:A\to B$
  
  
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*A relation that is univocal to the right is defined as: $\forall x \in A, \forall y,y':(xRy' \wedge xRy)\implies y'=y$.
  
*A relation that is total to the left is a relation that: $\forall x \in A, \exists y\in B: xRy$
  
  
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*Function is a relation that is univocal to the the right and total to the left.
  
  
  
  Given the definition at hand, we know that $f$ sends one $x\in C$ to one $f(x)\in D$ but as $f$ is not univocal to the left, sometimes $f^{-1}$ can send one $f(x)\in D$ to more than one $x$. And hence It's possible to send one $x \in C$ to $f(x)\in D$ such that $f^{-1}$ sends it back to $x\in C$ and other $x'\notin C$.

 A: Assume $f \colon A \to B$.
Proof. $(\Longrightarrow)$ Suppose $f(C) \subset D$. We want to show $C \subset f^{−1}(D)$, i.e., by definition of inverse image $f^{-1}(D) = \{x \in A : f(x) \in D \}$, we need to show: if $x \in C$, then $x$ satisfies $x \in A$ and $f(x) \in D$. Now, we suppose $x \in C$. Since $C \subset A$, we have $x \in A$. Since $x \in C$, we have $f(x) \subset f(C) \subset D$, by hypothesis $f(C) \subset D$. Thus $x \in A$ and $f(x) \in D$, i.e., $x \in f^{-1}(D) = \{x \in A : f(x) \in D \}$, and hence $C \subset f^{−1}(D)$.
$(\Longleftarrow)$ Similarly suppose $C \subset f^{−1}(D)$. We need to show $f(C) \subset D$, i.e., $f(x) \in D$ for every $x \in C$, by definition of image. Let $x \in C$. Then we have $x \in f^{-1}(D)$, by hypothesis $C \subset f^{−1}(D)$. Thus, by definition of inverse image, we have $f(x) \in D$, as desired.

Maybe, this exercise is important because the proof helps to see that the $\subset$ relation can not be improved to $=$. For instance, if $f \colon \mathbf{Z} \to \mathbf{Z}$ is the map $f(x) = x^2$, then $f^{-1}(\{0,1,4\}) = \{-2,-1,0,1,2\}$. But $f^{-1}(f(\{-1,0,1,2\})) \ne \{-1,0,1,2\}$.
