i have to prove the following statement:
$\mathbb \forall a \in A, \exists ! b \in B : (a,b) \in \Gamma $
where Gamma is the graph of a function. I proceeded in this direction (first time using absurd proof):
Lets say that a function is not a unique correspondence between elements of the domain and co-domain. Then, i can write:
$\mathbb \forall a \in A, \exists b \in B : (a,b) \in \Gamma $
This implies that there's atleast one correspondence. Atleast means that there can be $\ge$ 1 correspondence between a and b. If this is true, then the circle and the ellipse are functions (there are more drawings). If it is a function, then the following statement is true:
$\mathbb \Gamma = \{(a,b) \in AxB: b=f(a)\} $
which is the correct definition of a function. In the case of the ''functions'' like ellipse and the circle though, the definition falls since it changes to:
$\mathbb \Gamma = \{(a,b) \in AxB: b=f(a)=f(c)\} $
Not only this, but this last statement does not have the unique correspondance that defines a function.
Conclusion: if there's a graph of a function, then the correspondance between elements of domain and codomain is unique.