Just verifying if this proof is valid.
Claim: If $f:A \rightarrow B$ and $f^{-1}$ is a function, then $f$ is $1-1$.
Proof: Suppose $f(a)=f(z)$ where $a,z\in A$.
This implies that $(a,b)\in f$ and $(z,b)\in f$, where $b\in B$
Which then implies that $(b,a)\in f^{-1}$ and $(b,z)\in f^{-1}$
Since $f^{-1}$ is a function, $a=z$
So, $f(a)=f(b) \implies a=z$
Therefore $f$ is $1-1$
On
Daniel is right. Is not sufficient your proof.
Have good arguments but I think you can do better.
Take a look at this proof:
Let $a, b \in A$ and $f(a)=f(b)$.
Since, $f^{-1}$ is a function and $f(a), f(b) \in B$ with $f(a)=f(b)$,
we have that $f^{-1}(f(a))=f^{1}(f(b)) \rightarrow a=b$
Notice that I chose to prove 1-1 using: If $f(a)=f(b) \rightarrow a=b$.
You can also prove if the negation of this: If $a\neq b \rightarrow f(a) \neq f(b)$
It looks like a lot of the elements of a proper proof are there; however, your chain of reasoning is completely in the wrong order. In particular, just showing $\exists a\in A, \exists z\in A, f(a) = f(z) \wedge a = z$ is definitely not sufficient to show $f$ is one-to-one. What you need to show is: $\forall a\in A, \forall z \in A, f(a) = f(z) \rightarrow a = z$.