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The first answer to your question is "yes". In a topos, say, morphisms are in bijective correspondence with functional relations. More precisely, given objects $A$ and $B$, there is an isomorphism between the exponential $B^A$ and the object $FunRel(A,B)$ of functional relations between $A$ and $B$. In the internal language of the topos $FunRel(A,B)$ is described as (where $\exists!$ means "exists exactly one") $$FunRel(A,B) = \lbrace F \in \mathcal{P}(A \times B) \mid \forall x \in A \exists! y \in B . \; \langle x, y \rangle \in F \rbrace.$$ You will of course recognize this as the usual set-theoretic definition of a function as a collection of ordered pairs. The crucial bit of logic that is needed for the bijection is the principle of unique choice, which says that for every $R \subseteq A \times B$, $$(\forall x \in A \exists! y \in B . \; \langle x, y \rangle \in R) \implies \exists f \in B^A \forall x \in A . \langle x, f(x) \rangle \in R.$$
2. the definition of function as an element of a function type $\tau \to \sigma$ in a (simply typed) $\lambda$-calculus allows us to interpret functions as programs, see for example primitive recursive functionals. These do not in general satisfy the law of extensionality $(\forall x : \tau .\; f(x) = g(x)) \implies f = g$.