Is this a valid argument $f^{-1}[f(C)] = C$? I am given the fact that the function f is injective, and I want to prove the equivalence $f^{-1}[f(C)] = C$ 
My argument was:
"Suppose $x \in f^{-1}[f(C)]$. Since x $\in f^{-1}[f(C)]$, this means that $f(x) \in f(C)$. Now suppose that $z \in C$. Then we know that $f(z) \in f(C)$. Now suppose that $f(z)$ is the exact value that $f(x)$ is. Since we know that f is injective, then x = z, therefore $x \in C.$ Thus $f^{-1}[f(C)] \subseteq C$"
I know I need to prove the other direction, but is this argument sound? Thank you!
 A: You really should organize what you are doing. You start with $x$, do some stuff, then set it aside and introduce a new $z$, then come back to $x$... it’s a bit of a mess. 
Now, the idea seems to be fine, but the presentation isn’t. Let’s organize it a bit better:
To prove that $f^{-1}[f(C)]$, we take $x\in f^{-1}[f(C)]$; we want to show that $x\in C$. So we know that because $x\in f^{-1}[f(C)]$, then $f(x)\in f[C]$. That means that there exists $z\in C$ such that $f(z)=f(x)$ (because $y\in f[C]$ if and only if there exists $z\in C$ with $f(z)=y$; so apply this definition with $y=f(x)$). And because $f$ is injective, this means that $x=z$. That in turn tells us that $x\in C$ (because $x=z\in C$), which is what we wanted to prove.
That’s what you were doing, but organized as a single flow of arguments, instead of as two separate arguments that somehow meet somewhere in the middle of the paragraph.
The other direction, by the way, should not require the use of the hypothesis that $f$ is injective. The condition $X\subseteq f^{-1}[f(X)]$ hold for any function. 
