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According to the wikipedia on Affine Spaces, an Affine Space $A$ is defined as an underlying set $A$ alongside a vector space $\vec{A}$ with right group action (free, transitive) $+$ of $\vec{A}$ on $A$.

After laying out this definition, wikipedia states:

The uniqueness property allows us to define the subtraction of any two $a$ and $b$ of $A$, producing a vector of $$\overrightarrow{A}.$$ This vector, denoted

$$b - a,$$

or

$$ \overrightarrow{ab},$$

is the unique vector in $\overrightarrow{A}$ such that $ a + (b - a) = b$ (equivalently, $a+\overrightarrow{ab}=b$).

This subtraction has the two following properties, called Hermann Weyl's axioms.

  1. $ \forall a \in A,\; \forall v\in \overrightarrow{A},$ there is a unique point $ b \in A$ such that $ b-a = v$ (equivalently, $\overrightarrow{ab}=v$), and
  2. $ \forall a,b,c \in A,\; (c - b)+ (b - a) = c - a$ (equivalently, $\overrightarrow{ab} + \overrightarrow{bc} = \overrightarrow{ac}$).

Question: Is the "uniqueness property" an additional axiom that Affine Spaces must satisfy, or does it somehow follow from its definition?

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user1770201
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1 Answers1

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The uniqueness property follows from the fact that for any $a \in A$, the map $\vec{A} \to A$, $v \mapsto a + v$ is a bijection. If $a, b \in A$, the vector $\vec{ab}$ is defined as the inverse image of $b$ under such function.

Luca Bressan
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  • How do you know that that map is injective (I understand why it is surjective)? You might say "because the group action is defined to be free", but that only shows that $v \not \mapsto a$ unless $v = 0$. – user1770201 Sep 09 '16 at 22:37
  • @user1770201 suppose a+v=b and a+w=b. Then a+v=a+w. Thus, a+v-w=a+w-w=a+0=a. Thus, v-w=0. – ASKASK Sep 09 '16 at 22:58
  • @user1770201: if you're using the axioms given in the Wikipedia article, you know that the map $\vec{A} \to A$, $v \mapsto a + v$ is bijective, not just surjective. In fact, if the map were only surjective, there could be many ways to choose $\vec{ab}$ from the inverse image of $b$. Once we know it is also injective, there is exactly one vector in the inverse image of $b$, that we call $\vec{ab}$. Then the map $A \to \vec{A}$, $b \to \vec{ab}$ is bijective as well, since it is the inverse of a bijection. – Luca Bressan Sep 10 '16 at 07:09