Prove that addition of a constant on vector spaces is bijective

Question

What would be a nice way to deduce from the vector space axioms that $f : V_1 \longrightarrow V_2, \, x\mapsto x+v$ with constant $v$ is bijective?

Answer 1

To prove that what you've described is a bijection, you have to show that it's one-to-one and onto.

To show that it's one-to-one, we let $x$ and $y$ be vectors that both map to the same vector $x + v$, and we want to show that $x$ and $y$ must be equal.

Since $y \longrightarrow y + v$, we have that $x + v = y + v$. It follows from the inverse axiom that $v$ has an additive inverse $-v$, and that

$x + v + -v = y + v + -v$,

$x + 0 = y + 0$,

and $x = y$ by the identity element axiom.

To show that it's onto, we can show that every vector in our space is included in the image of our function, ie. it can be expressed as $x + v$.

So suppose the vector $x$ is in our range. Then

$x = x + 0$ by the identity of addition axiom,

$x = x + -v + v$ by the additive inverse axiom, and

$x = (x + -v) + v$ by the associativity axiom.

Therefore there is a vector in $V_1$ that maps to $x$, namely $(x + -v)$.

Answer 2

It is onto since given any $y \in V_2$, we have $y-v \mapsto y$.

It is 1-1 since $x+v = y+v \implies x=y$.

Another way is to note that the inverse map is $x \mapsto x-v$.