Is it true that $f: S\to S$ be a function: $(f \circ f)$ is bijejective if, and only if $f$ is bijective? 
Let $f
 :S\rightarrow S$
    be a function.  Show that $f\circ f$
    is bijective if, and only if, $f$
    is bijective.

My solution.
If $f\circ f$
  is bijective if, and only if, $f$
  is bijective and $f$
  is bijective if, and only if $f\circ f$
  is bijective. Suppose $f\circ f
  :S\rightarrow S$
  is bijective if and only if $f$
  is bijective. We know that $f\circ f$
  is a fuction. Suppose $x = y$, then we know that :
$$f\circ f
  (x) = f\circ f
  (y)$$
$$f(f(x)) = f(f(y))$$
This implies that $f(x) = f(y)$. Hence $f$ is a function.


*

*We know that $f\circ f$
is injective. Suppose $f(x)=f(y)$. Then applying $f$ again we have $f(f(x)) = f(f(y)) \Longrightarrow
 f\circ f
  (x) = f\circ f
  (y) \Longrightarrow
 x=y$. Hence $f$ is injective.

*Let $y\in
 S$. We know that $f\circ f$
is surjective so there exists an $x \in
 S$. such that $f\circ f
  (x) = y$ and $f(f(x)) = y$. Since $f$ is mapped from $S$ to $S$, then $f(x)\in
 S$. 
Denote $f(x) = z$, then $f(z) = y$. Hence $f$ is surjective.
So $f$ is a function, injective and surjective, thus $f$ is bijective which implies that $f\circ f$
  is bijective.
Can anyone provide me with some feedback?
 A: Hint: Let $f\colon A\to B$, $g\colon B\to C$ be functions. If $g\circ f$ is injective then $f$ is injective. If $g\circ f$ is surjective then $g$ is surjective. 
A: *

*Injective: $f: S\hookrightarrow S \iff \Big(\forall x,y\in S: f(a)=f(b) \implies a=b\Big)$
$$\begin{align}\because f: S\hookrightarrow S & \iff \Big(\forall f(x),f(y)\in S: f(f(x))=f(f(y)) \implies f(x)=f(y)\Big) \\ & \iff \Big(\forall x,y \in S: f\circ f(x)=f\circ f(y)\implies x=y\Big) \\ \therefore f: S\hookrightarrow S &\iff f\circ f : S\hookrightarrow S & \end{align}$$


*Surjective: $f: S\twoheadrightarrow S \iff \Big(\forall x \in S, \exists y\in S: f(y)=x \Big)$
$$\begin{align}\because f: S\twoheadrightarrow S & \iff \Big(\forall f(y)\in S, \exists f(z) \in F : f(f(z))=f(y)\Big)\\ & \iff \Big(\forall x\in S, \exists z \in S: f\circ f(z)= x\Big) \\\therefore f: S\twoheadrightarrow S & \iff f\circ f: S\twoheadrightarrow S\end{align}$$

*

*Bijective: $f:S \leftrightarrow S \iff f:S \hookrightarrow S \bigwedge f:S \twoheadrightarrow S$
$$\therefore f:S\leftrightarrow S \iff f\circ f: S \leftrightarrow S$$
A: Hint (different approach than Hagen): f is bijective if and only if it is invertible.
A: Since you explicitly asked for feedback, here you go: I rewrote your answer and made some remarks.
Let $f: S \to S$ be a function. We prove that $f \circ f$ is bijective if, and only if, $f$ is.
Firstly, suppose that $f \circ f$ is bijective. We prove that $f$ is then also bijective.
[Why do you show that $f$ is a function? The first assumption you made is that $f$ IS a function. Furthermore, could you elaborate on why $f\big(f(x)\big) = f\big(f(y)\big) \implies f(x) = f(y)$?]
We know that $f\circ f$ is injective. Suppose $f(x) = f(y)$. Then applying $f$ again we have $$f\big(f(x)\big) = f\big(f(y)\big) \implies f \circ f(x) = f\circ f(y) \implies x = y \, .$$ Hence $f$ is injective.
Now let $y\in S$. We know that $f \circ f$ is surjective so there exists an $x\in S$ such that $f \circ f(x) = y$ and $f\big(f(x)\big) = y$. Since $f$ is mapped from $S$ to $S$, we have $f(x) \in S$. Denoting $f(x) = z$, we have $f(z) = y$. Hence $f$ is surjective.
We now conclude that $f$ is bijective if $f \circ f$ is bijective.
So far so good for the proof of $f \circ f$ is bijective $"\implies"$ $f$ is bijective.
To do list: show the other direction $f$ is bijective $"\implies"$ $f \circ f$ is bijective.
A: 
For a function $f:S\to S$ be a function: $f\circ f$ is bijective if, and only if, $f$ is bijective.
MY SOLUTION:
If $f\circ f$ is bijective if, and only if, $f$ is bijective and $f$ is bijective if, and only if $f\circ f$ is bijective. 

This appears to be what you are trying to demonstrate.  It is also redundant.  "If and only if" is symmetric.  Perhaps you should say something like:
"We shall show that $f\circ f$ is bijective if $f$ is bijective, and that $f$ is bijective if $f\circ f$ is bijective."

Suppose $f\circ f:S\to S$ is bijective if and only if $f$ is bijective. We know that $f\circ f$ is a function. 

Don't assume your conclusion unless you are attempting to demonstrate a contradiction.

Suppose $x=y$, then we know that :
$$f\circ f(x)=f\circ f(y)$$
$$f(f(x))=f(f(y))$$
This implies that $f(x)=f(y)$. Hence $f$ is a function.

How is that implied?
What you need to show here is that if $f$ is injective then it follows that $f\circ f$ is too.


*

*If $f$ is injective then, by definition:

*

*$\forall a,b\in S, f(a)=f(b)\implies a=b$.  


*By substitution, $f(x)=a, f(y)=b$, then

*

*$\forall f(x),f(y)\in S, f(f(x))=f(f(y))\implies f(x)=f(y)$


*Now, since $f:S\to S$ then $\forall x\in S, f(x)\in S$ so:

*

*$\forall x,y\in S, f(f(x))=f(f(y))\implies f(x)=f(y)$


*And if $f(x)=f(y)\implies x=y$ then:

*

*$\forall x,y\in S, f(f(x))=f(f(y))\implies x=y$


*Which is what we needed to show.



We know that $f\circ f$ is injective. Suppose $f(x)=f(y)$. Then applying $f$ again we have $f(f(x))=f(f(y))\implies f\circ f(x)=f\circ f(y)\implies x=y$. Hence $f$ is injective.

We don't know that $f\circ f$ is injective.  But you can show that if it is, then f has to be too.

Let $y\in S$. We know that $f\circ f$ is surjective so there exists an $x\in S$. such that $f\circ f(x)=y$ and $f(f(x))=y$. Since $f$ is mapped from S to S, then $f(x)\in S$.
Denote $f(x)=z$, then $f(z)=y$. Hence f is surjective.

Again we don't know that $f\circ f$ is surjective. But you have shown that if it is then $f$ must be. 

So $f$ is a function, injective and surjective, thus $f$ is bijective which implies that $f\circ f$ is bijective.

You seem to have missed the step where you show this to be so.

CAN ANYONE PLEASE ME FEEDBACK ON MY SOLUTION

Overall you do seem to know what you are trying to do.  You just don't seem to be able to express it.  Work on how you phrase things.  (Mostly, if you assume something you don't "know" it.)
