If $f:A \to B$ is a bijection, show that $g:B \to A$ is a function. So, a textbook I'm reading from defines a function as follows: 

A function $f$ from a set $A$ to a set $B$ is the set
  $$\Gamma_f := \{(x,y)\in A \times B : y = f(x)\}$$ with the property that $\forall x \in A, \exists! y \in B$ such that $(x,y) \in \Gamma_f$

Now, I'm told to prove the following:
Let $f: A \to B$ be a bijection. Define $g: B \to A$ such that $g(y) = x$ precisely when $f(x) = y$. Show that $g$ is a function.
Proof Attempt: First, we wish to show that $$\Gamma_g = \{(y,x) \in B \times A: x = g(y)\}$$ is non-empty. Let $b \in B$. Since $f$ is surjective, there exists an element $a \in A$ such that $f(a) = b$. By the definition of $g$, we have $g(b) = a \implies (b,a) \in \Gamma_g$, so $\Gamma_g \neq \emptyset$. 
Now, to show the uniqueness property, suppose $(b,a') \in \Gamma_g$. Then, by the definition of $\Gamma_g$, $a' = g(b)$ which means that $f(a') = b$. So, $f(a') = f(a)$. By the injectivity of $f$, it follows that $a' = a$. Hence, $(b,a') = (b,a)$. This completes the proof.
Is my proof correct? Did I show the right things? If not, what should I have shown and what is the correct way path I should have taken? Any advice would be appreciated. 
 A: In the first part it is not enough to say i that $\Gamma_g$ is non empty.  What you need is: for every $b\in B$ there exists $a\in A$ such that $(b,a) \in  \Gamma_g$. You have proved this but you have not stated the argument correctly.
A: Looks good to me, except that in order for $g$ to be a function, $\forall x\in B,\exists a\in A$ such that $g(b)=a$, that is, every $b\in B$ has an image in $A$. You have proved this in the first part of the proof by invoking the surjectivity of $f$, but labelled it as an attempt to show that $\Gamma_g$ is non-empty.
A: Some critics not on your question but on those who gave this to you as a question.
Too much for a comment.

"Define $g:B\to A$ such that $g(y)=x$ precisely when $f(x)=y$."
You were told to do that (so cannot be blamed for it) but this looks like a false start to me.
The notation $g(y)=x$ already indicates that $g$ is a function, and this "definition" of $g$ is due to the injectivity of $f$ (what should at least have been mentioned).
They should have asked you to prove that relation $$(\Gamma_f)^{-1}:=\{(y,x)\mid y=f(x)\}\subseteq B\times A$$is a function.
After that (not before) this function can be labeled with the name $g$ and expressions like $g(y)=x$ become meaningful.

A second thing: "A function $f$ from a set $A$ to a set $B$ is the set:$$\Gamma_f:=\{(x,y)\mid y=f(x)\}\subseteq B\times A$$with the property that....
There they seem to say that $f=\Gamma_f$ and as a consequence that $f$ is defined by means of $f$...
