Proving that $C$ is a subset of $f^{-1}[f(C)]$ More homework help.  Given the function $f:A \to B$.  Let $C$ be a subset of $A$ and let $D$ be a subset of $B$.
Prove that:
$C$ is a subset of $f^{-1}[f(C)]$
So I have to show that every element of $C$ is in the set $f^{-1}[f(C)]$
I know that $f(C)$ is the image of $C$ in $B$ and that $f^{-1}[f(C)]$ is the pre-image of $f(C)$ into $A$.  Where I'm stuck is how to use all of this information to show/prove that $C$ is indeed a subset.  
Do I start with an arbitrary element (hey, let's call it $x$) of $C$? and then show that $f^{-1}[f(x)]$ is $x$?  I could use a little direction here...  Thanks.
 A: Copied from my answer to I am issues with proving the following problem: $f^{-1}(f(A)) ⊃ A$, which was closed as a duplicate of this question just before I posted it.

$$
f^{-1}\left(f(A)\right)=\left\{x:f(x)\in f(A)\right\}\tag{1}
$$
Note that if $x\in A$, then $f(x)\in f(A)$, and by $(1)$, $x\in f^{-1}\left(f(A)\right)$. Therefore, by definition, we have
$$
A\subset f^{-1}\left(f(A)\right)\tag{2}
$$
However, if $f$ is not injective, then $f^{-1}\left(f(A)\right)$ may indeed contain elements not present in $A$; for example let $f:\mathbb{Z}\mapsto\mathbb{Z}$ be defined by
$$
f(x)=\left\lfloor\frac x2\right\rfloor\tag{3}
$$
and let $A$ be the set of even integers. Then $f(A)=\mathbb{Z}$ and 
$$
A\subsetneq\mathbb{Z}=f^{-1}\left(f(A)\right)\tag{4}
$$
A: Since you want to show that $C\subseteq f^{-1}\big[f[C]\big]$, yes, you should start with an arbitrary $x\in C$ and try to show that $x\in f^{-1}\big[f[C]\big]$. You cannot reasonably hope to show that $f^{-1}\big[f[\{x\}]\big]=x$, however: there’s no reason to think that $f$ is $1$-$1$, so there may be many points in $A$ that $f$ sends to the place it sends $x$.
Let $x\in C$ be arbitrary. For convenience let $E=f[C]\subseteq B$. Now what elements of $A$ belong to the set $f^{-1}\big[f[C]\big]=f^{-1}[E]$? By definition $f^{-1}[E]=\{a\in A:f(a)\in E\}$. Is it true that $f(x)\in E$? If so, $x\in f^{-1}[E]=f^{-1}\big[f[C]\big]$, and you’ll have shown that $C\subseteq f^{-1}\big[f[C]\big]$.
A: The problem can be reduced to one-line proof in the following way.
Let $f : Y \leftarrow X$ denote a function.
The definition of inverse images says that for all $y : Y$ and all $x : X,$ we have:
$$f^*(y) \ni x \iff y = f(x)$$
This can be rewritten:
$$f^*(y) \supseteq \{x\} \iff \{y\} \supseteq \{f(x)\}$$
This can be used to prove:

Proposition. For all subsets $B$ of $Y$ and all subsets $A$ of $X$, we have:
$$f^*(B) \supseteq A \iff B \supseteq f_*(A)$$

Once you've proved this, your problem becomes a one-line proof:
$$f^{*}(f_*(A)) \supseteq A \iff f_*(A) \supseteq f_*(A)$$
