Let $A = \{1\}, B = \{3\}$ Suppose $f = \{(1,3),(5,6)\}$ Is $f: A \rightarrow B$? Let $A = \{1\}, B = \{3\}$ 
  
  Suppose $f = \{(1,3),(5,6)\}$ Is $f$ a function from $A$ to $B$?


The definition of the function, say $F$, that the book gives:

$$\tag 1 \forall a \in A \exists! b \in B ((a,b) \in F)$$

Then $F: A \rightarrow B$

Coming back to our example: although we have $(5,6) \in f$ and $5 \notin A$ and $6 \notin B$, but, if I get it correctly, main condition of the function is not violated, hence I believe that yes, $f$ is a function from $A$ to $B$. 

I've tried to answer my own question, but I need your verification:  is $f$ a function from $A$ to $B$?
Usually a function $f$ is from $A$ to $B$, we have $$f\subseteq A\times B,$$
so I would say that $f$ is not $A\to B$. 

If that's your only definition, then technically it meets it, but it's not standard.
The domain and range of $f$ are not $A$ and $B$ respectively, but instead are $\{1,5\}$ and $\{3,6\}$ respectively.