Predicate logic for statements about functions? Given the relation $R$ defined on the cartesian map $A\times A$ where $|A|=n$. 
How to use the predicate logic to express the statements about functions? 
Examples.


*

*The relation R corresponds to a function from A to A. 

*The relation R corresponds to an injective function from A to A.

*The relation R corresponds to and bijective function from A to A.


where for each part above respectively I think them as


*

*$\forall a\in A$, there exist $c\in A,s.t.(a,b)\in R$.

*$\forall a,b\in A$, there exist $c,d\in A,s.t. (a,c),(b,d)\in R\& (a\ne b\to c\ne d)$

*i8t is surjective with the statement in ii). 


Are the above interpretations right?
 A: Let $=$ be the equality predicate. Then the relation $R$ is a function if
$$
\forall x \forall y \forall z: ((x, y) \in R \land (x, z) \in R) \to (y = z).
$$
It is a total function if it is a function and
$$
\forall x \exists y: x \in A \to (x, y) \in R.
$$
Usually we omit "total" and assume that all "functions" are total. (Those that are not total are given the adjective "partial".) So I believe you need both statements in your problem.
In order for $R$ to be an injective function, it must be a (total) function and satisfy
$$
\forall x \forall y \forall z: ((x, z) \in R \land (y, z) \in R) \to (x = y).
$$
Finally, in order for $R$ to be a bijective function, it has to be injective and satisfy
$$
\forall y \exists x: y \in A \to (x, y) \in R.
$$
If your quantifiers are always over $A$, predicates for surjectivity and totality can be shortened a little bit. Here's the summary:


*

*Well-defined: $\forall x \forall y \forall z: ((x, y) \in R \land (x, z) \in R) \to (y = z)$

*Total: $\forall x \exists y: (x, y) \in R$

*Injective: $\forall x \forall y \forall z: ((x, z) \in R \land (y, z) \in R) \to (x = y)$

*Surjective: $\forall y \exists x: (x, y) \in R$

*Being a (total) function: Total and well-defined

*Being an injective (total) function: Total, well-defined and injective

*Being a bijection: Total, well-defined, injective and surjective

A: Not exactly.
i) $\phi:=$"$\forall a\,\exists b:aRb\land \,\big(\forall c: (aRc\Rightarrow b=c)\big)$" 
ii) $\psi:=\phi\land$ "$\forall a,b,c: (aRc\land bRc)\Rightarrow (a=b)$" 
iii) $\vartheta:=\psi\land$ "$\forall b\,\exists a: aRb $"
A: No, they aren't:
The first statement you wrote would be satisfied if the set $A$ were ${1,2}$ and $R$ were defined as ${(1,1),(1,2),(2,2)}$. That is, you need to require also that $b$ is unique.
For the second statement, I think you almost have it but I've usually seen it phrased the other way, that is you start by asserting that $R$ is a function, (using what you said for the first statement) and then say that $c=d \to a=b$. But if you fix your first statement, your way works too.
For the third statement, yes, I'd just say the second statement and surjective (which is just $\forall b \in A, \exists a \in A {\rm s.t.} (a,b) \in R$ )
