Affine space definition I am studying affine space
But when i see its definition on YouTube it looks like
Let $A$ is set of points and $V$ is vector space over R and $f:A×A\to V$ such that
$\forall$ $x, y, z\in A $
$f(x,y)+f(y,z)=f(x,z)$ and $\forall$ $p$ $f_{p}:A\to V$ such that $f_{p}(x)$=$f(p, x) $ is $1-1$ then it is affine.
But i downloaded a book where definition is,
An affine space over field $K$ is a triple $(A, L, +) $ consisting of a linear space $L$ over a field $K$, a set $A$ whose elements are called points, and an external binary operation $A×L\to A$ satisfying the following axioms.
$(a+l)+m=a+(l+m)$. $a+0=a$. for any two points $a, b\in A$ $\exists$ a unique vector $l$ with the property $b=a+l$
Please explain what are the relationship between them?
 A: I'm only at all familiar with the former. The function $f$ is, in a sense, "subtraction". It gives us a way to "subtract" two points in the affine space, to obtain the relative vector between the two points. For example, if we consider points $v, w$ in  the affine space $\Bbb{R}^n$, then $f(v, w)$ is the vector $\langle w_1 - v_1, w_2 - v_2, \ldots, w_n - v_n\rangle$. Note that it satisfies the conditions required.
In this way, we can talk about points relative to each other, but not in any absolute terms. The $0$ vector in $V$ gives us an absolute frame of reference, but when considered in an affine space, the $0$ vector simply refers to the unique vector between two of the same points.
The invertibility of $f_p$ implies that there is an inverse: there should be a way to "add" a vector in $V$ to a point in the affine space $A$. Given any $y \in A$, there is a unique $v \in V$ such that $f(p, y) = v$, so we can, in a sense, say that $y$ is $p$ with the vector $v$ added on. We can even define an addition operation $+ : A \times V \to A$ with this in mind.
This takes us to the next definition. Instead of starting with a vector-valued subtraction of points in $A$, they start with the inverse operation: a point-valued addition of vectors and points. Note that they still assume some invertibility of the map $\ell \mapsto a + \ell$, which allows you to form a subtraction operation as per the first definition.
I won't verify that these conditions are indeed equivalent; I'll leave that to you. But this is how they interconnect.
A: Let $(A,V,f)$ be an affine space (1st definition). $V$ is a vector space over a field $K$.
Let us define $A\times V\to A, (a,l) \to f_a^{-1}(l)$, what we will denote $a+l$.

*

*$\forall a,b \in A, a+l=b \Leftrightarrow  f_a^{-1}(l)=b \Leftrightarrow  l=f_a(b)=f(a,b)$. So, $\exists !l : a+l=b$;

*$\forall a \in A, a+0=f_a^{-1}(0)=a$ because $f(a,a)+f(a,a)=f(a,a)\Rightarrow f(a,a)=0$;

*$\forall a \in A, \forall l \in V, l=f(a,a+l)$ and $f(a+l,(a+l)+m)=m$ and $f(a,a+l)+f(a+l,(a+l)+m)=f(a,(a+l)+m)$. So, $f_a^{-1}(l+m)=(a+l)+m$. So, $(a+l)+m=a+(l+m)$.

So, A is an affine space(2nd definition).
In the other direction, it is the same type of reasoning that becomes clear when we rely on drawings (after all, it is geometry...)
