Addition between topological vector spaces is bijective?

Question

Let $X$ be a topological vector space. Then it is said that for any $x_0 \in X$ the mapping $x\rightarrow x + x_0$ is bijective (in fact a homeomorphism). In the proof there is only written that it follows from the vector space axioms. However, addition in general is not bijective since $1+3=2+2 $ for instance. So why is it different here?

Answer 1

Your map only has one variable, your addition has two. If you let $x_0$ to be fixed, then the map $x \mapsto x_0 + x$ is bijective but the map $(x, y) \mapsto x+y$ is not, because $0+0 = v + (-v)$ with $v$ a nonzero vector.

For a proof that it is bijective, try finding explicitly the inverse map.

Answer 2

$T_{x_0} : X\to X$ defined by $T_{x_0}=x_0+x$

1. $T_{x_0}(x) =T_{x_0}(y)$ implies $x_0+x=x_0+y$ implies $-x_0+(x_0+x) =-x_0+(x_0+y) $

Use associativity of vector addition and $-x_0$ as additive inverse of $x_0$ to conclude $x=y$

2. Let $y\in X$ then $\exists x(=-x_0+y)\in X$ such that

$\begin{align}T_{x_0}(x) &=T_{x_0}(-x_0+y)\\&=x_0+(-x_0+y) \end{align}$

Vector addition is associative and $-x_0$ is additive inverse of $x_0$. Then conclude $T_{x_0}(x) =y$

Note: First you need to see the difference between addition map and translation map ( i.e when one coordinate of addition map is fixed)