Peano axioms and meaning of successor map in Jacobson's Basic Algebra I

Question

From Jacobson's Basic Algebra I on P. 16, the Peano axioms are stated as:

1. $0 \neq a^+$ for any $a$ (that is, $0$ is not in the image of $\mathbb{N}$ under $a \to a^+$).
2. $a\to a^{+}$ is injective.
3. (Axiom of induction). Any subset of $\mathbb{N}$ which contains $0$ and contains the successor of every element in the given subset coincides with $\mathbb{N}$.

First, $a$ does not seem to be defined to be anything (at least from what I can see on the current and previous page), though it seems like it is supposed to be representing some arbitrary element in $\mathbb{N}$? It is not clear. So my first question is what is the notation $a$ and $a^+$ referencing?

The other question I have is regarding that of the successor map. Why do I need two axioms for a map? Why can I not just point to the mapping $f: \mathbb{N} \to \mathbb{N}$ given by $f(n) = n+1$. When I think more about it, couldn't I define the map to be $f(n) = n+k$, where $k$ is a positive integer? What is the relation of the axiomatization to the successor map? Does the successor map not exist without an axiomatization?

Additional Context

There are only two other places where $a$ is mentioned before this, which is on P.15 of the text: "For example, in a structure $S$ in which an associative binary composition and a unit are defined, any element $a \in S$ defines a map $n \to a^n$ where $a^0=1, a^1 = a,$ and $a^k = a^{k-1}a$."

"From this point of view, we begin with a non-vacuous set $\mathbb{N}$, a particular element of $\mathbb{N}$ designated as $0$ and a map $a \to a^+$ of $\mathbb{N}$ to itself, called the successor map."

Answer 1

The Peano axioms are a way to define the natural numbers; to introduce the concepts of addition when trying to define the natural numbers would be circular.

So: "for any $a$" means "for any $a$ in $\bf N$". It is not assumed that $\bf N$ has any structure or properties other than those we can prove from the Peano axioms. After we have proved a few theorems, we can identify $a^+$ with $a+1$, but when we're just getting started, we don't have any "$+$" and we don't have any "$1$".