Suppose we have a function $f: A\to B$. Then we know, without specifying what $f$ is, that $f$ may or may not map to every element $b\in B$. If $f$ does map to every element $b\in B$ then it's surjective.
However, does the notation $f:A\to B$ imply that every element $a\in A$ is mapped to some unique $b\in B$?
Furthermore, is there a difference between domain and natural domain? If there is, how can we tell whether $A$ is a/the natural domain from the notation $f: A\to B$?
Suppose we consider the function defined this way: $f: \mathbb{R}\to \mathbb{R}, f=\frac{1}{x}$ Clearly, $f$ is not defined at $x=0$. Would it therefore be incorrect to write $f: \mathbb{R}\to \mathbb{R}, f=\frac{1}{x}$ instead of $f: \mathbb{R}\backslash \{0\}\to \mathbb{R}, f=\frac{1}{x}$?
The question is a bit clumsy. You are welcome to ask for clarification if this isn't clear enough.
In general, the notation $f:A \to B$ implies that every element $a \in A$ is mapped to some element $b∈B$, which is not necessarily unique.
However, the definition of a function means that this element is unique in terms of a, ie. for any $a\in A$ there exists only one $b \in B$ such that $f(a)=b$. If you are requiring uniqueness in terms of b as well, ie. for any $b \in B$ there can exist only one $a \in A$ such that $f(a)=b$, then the mapping is said to be injective. To write this formally, you could state:
\begin{align} f(x)=f(x') \Rightarrow x = x' \end{align}
This does not necessarily mean that every element of $B$ does have a corresponing element from $A$; It is very easily to see that the domain {1,2} can map injectively to {1,2,3} with each number in $A$ mapping to itself in $B$. This would meet the criteria for being an injective function, but would not be surjective.
Regarding domain vs. natural domain, I believe that the natural domain is the largest domain for which the mapping is defined. Every other domain must necessarily be a subset of the natural domain.
In your example, $A$ would be the natural domain as $f$ maps every single element in $A$ to a corresponding element in $B$, and any other domain will undoubtedly be a subset of it - the function is not specified to map elements outside of $A$ because we don't know what $A$ actually is.
As a counter-example, if we knew that $f: \mathbb R \to \mathbb R$, and were then given $f: A\to B$ with $A \subset \mathbb R$, then A would not be the natural domain any longer, as we would be aware that there exists a larger domain for which the function operates, ie. $\mathbb R$.