Prove that RE is closed under reduction Prove that the class RE is closed under reduction.
Definitions: 
A language $ A \subseteq \Sigma^*$ is called reducible to $ B \subseteq \Gamma^*$ ( denoted by $A \leq B$) if there is a computable function $ f: \Sigma^* \rightarrow \Gamma^*$ such that $\forall x \in \Sigma^*: x \in A \iff f(x) \in B$.
A class C is closed under $\leq$, if for all languages A, B it follows that: $ A \leq B \wedge B \in C \Rightarrow A \in C$.
A language A is recursively enumerable, if either A is empty or A is the image im(f) of a computable function $f: \Gamma^* \rightarrow \Sigma^*$ for an arbitrary $\Gamma$.
This is my attempt at proving that RE is closed under $\leq$.
Assume $A \leq B$ and let $f: \Gamma^* \rightarrow \Sigma^*$ be a function such that $ B = im(f)$.
Therefore there is a function for which all $f(x)$ are in B. Since $A\leq B$ it follows that there is a function such that $x \in A \iff f(x) \in B$. Let f be that function, so that A is the domain of that function. If you take the inverse function $f^{-1}$, you can recursively enumerate all $x \in A$, therefore $A \in RE$ and thus RE is closed under $\leq$.
Now a couple of questions. First is this proof correct? Secondly if a function is supposed to map from $\Gamma^*$ to $\Sigma^*$ can I simply assume that $A \subseteq \Gamma*$? 
And since f has to be computable do I have to say which elements f maps to if $x \in \Gamma^* \setminus A$?
 A: No the proof is not correct. The function witnessing $A\leq B$ and the one witnessing $B\in RE$ have no reason to be the same. Also none of these two functions have any reason to be inversible (so your $f^{-1}$ does not exist).
Here is a proof:
Assume $A\leq B$, witnessed by a computable function $f:\Sigma_A^*\to\Sigma_B^*$ with $x\in A\Leftrightarrow f(x)\in B$. Assume also $B\in RE$, witnessed by $B=Im(g)$ for some computable $g:\Sigma^*\to\Sigma_B^*$. We now build the computable function $h:\Sigma_A^*\times\Sigma^*\to\Sigma_A^*$, by defining $$h(u,v)=\left\{\begin{array}{l}u\text{ if }f(u)=g(v)\\
\text{undefined otherwise}
\end{array}\right.$$
If you don't like "undefined" you can replace it by any constant word $u_A\in A$ as soon as $A\neq\emptyset$ (if $A=\emptyset$ the result is trivial).
Now, we can prove that $Im(h)=A$ and therefore $A\in RE$. If $u\in A$, then $f(u)\in B$, and so there is some $v$ such that $f(u)=g(v)$, so we have $u=h(u,v)$ and $u\in Im(h)$.
Conversely, if $u\in Im(h)$, then there is $v$ with $f(u)=g(v)\in B$ and therefore $u\in A$. We proved $Im(h)=A$.
