Let F denote the set of all functions from $\{1,2,3\}$ to $\{1,2,3\}$, how do I prove or disprove the following:? My proofs look and sound stupid and I am sure they are all wrong.
Please help me figure this out.    

$a)$ One of $i)$ and $ii)$ is true and one is false.

$i)$ $\forall f \in F,$ $\exists g \in F$ so that $g(f(1)) = 2$.
My solution:
Let $f(1) = n$, $n \in \{1,2,3\}$. Then $g(n) = 2$.
$ii)$ $\forall f \in F$, $\exists g \in F$ so that $f(g(1)) = 2$.
Negation: $\exists f \in F$, $\forall g \in F$ so that $f(g(1)) \neq 2$.
My solution:
Let $g(1) = n$, $n \notin \{1,2,3\}$. Then $g(n) \neq 2$.

$b)$ Let $f \in F$ be defined by $f(1)=2$, $f(2)=3$, $f(3)=2$. Find and simplify the number of functions $g\in F$ so that $f(g(f(1))) =2$.

My solution:
There are three functions $g$, such that $g(n)$, $n \in \{1,2,3\}$, $g(n)=1$ or $2$ or $3$.
 A: For (a) you’ve correctly determined which is true and which is false, but your argument for $(i)$ is incomplete, and your argument for $(ii)$ is incorrect. 
For $(i)$ you need to show that no matter what $f$ is (so long as it’s in $F$, of course), there is some $g\in F$ such that $g\big(f(1)\big)=2$. To do this, you must explain how to construct all of $g$. You’ve done the most critical part by saying that if $f(1)=n$, then we want to be sure that $g(n)=2$, but you still have to say what the other two values of $g$ are. Fortunately, it doesn’t matter: no matter what they are, it will be true that $g\big(f(1)\big)=g(n)=2$. The simplest solution is to make $g$ the constant function that sends every element of $\{1,2,3\}$ to $2$, but it’s certainly not the only choice; we could just as well let define $g$ by
$$g(k)=\begin{cases}
2,&\text{if }k=n\\
1,&\text{if }k\ne n\;,
\end{cases}$$
for instance.
For $(ii)$ you need to show that there is at least one $f\in F$ such that $f\big(g(1)\big)\ne 2$ no matter which function from $\{1,2,3\}$ to $\{1,2,3\}$ you choose for $g$. In particular, this means that you have to find a specific function $f\in F$ that has a certain property with respect to every $g\in F$; you don’t get to specify $g$. Moreover, all of the functions $f$ and $g$ under discussion here are functions in $F$, i.e., from $\{1,2,3\}$ to $\{1,2,3\}$, so a function $g$ such that $g(1)\notin\{1,2,3\}$ is completely irrelevant to the problem: it’s not in $F$. The key here is to notice that no matter what function $g$ is, $f\big(g(1)\big)$ is $f(n)$ for some $n\in\{1,2,3\}$. Thus, if we choose an $f$ such that $f(k)$ is never $2$, it won’t matter what $g$ is: $f\big(g(1)\big)$ won’t be $2$. There are $2^3$ functions $f\in F$ such that $f(k)$is never $2$; can you find one of them?
For (b) you know that $f(1)=2$, so you want all of the functions $g\in F$ such that $f\big(g(2)\big)=2$. 


*

*If $f\big(g(2)\big)=2$, what are the possible values of $g(2)$?  

*And what are the possible values of $g(1)$? Of $g(3)$?


Now put the pieces together as we did in your earlier question to get the correct answer to (b); it’s quite a bit more than $3$.
A: (i) You have the right idea, but you need to define a function $g$, not just give a single value.  For example, you could define $g(1)=g(2)=g(3)=2$, and then you no longer care about what $f$ does to 1.
(ii) This solution is incorrect.  You need to find an $f$ that works for all $g$.  Try $f(1)=f(2)=f(3)=1$.
(b) This solution is incorrect.  You need to choose $g$ so that $g(2)\neq 2$, but $g$ is arbitrary otherwise.
