Show that $\forall x\in \Bbb{R}$ and for a fixed $a\in \Bbb{R},\;t_a= a+x$ is invertible

Question

Let $a\in\Bbb{R}$ be fixed. Define \begin{align} t_a:\,&\Bbb{R}\to \Bbb{R}\\ &x\mapsto a+x\end{align} I want to show that $t_a$ is invertible. Below is my trial

MY WORK

Suppose $t_a v=0$, then $av+xv=0\implies v(a+x)=0\implies v=0,$ and we are done.

Please, I'm I right? If no, can anyone help out?

Answer 1

Here inverse should be taken w.r.t. composition of function i.e. you have to find another function $L:\Bbb R\rightarrow \Bbb R$ , such that $t_a\circ L=L\circ t_a= Id_{\Bbb R}$. Notice that in this case $t_{-a}:\Bbb R\rightarrow \Bbb R$ defined by $$\begin{align} t_{-a}:\,&\Bbb{R}\to \Bbb{R}\\ &x\mapsto -a+x\end{align}$$ do the job i.e. $$t_a\circ t_{-a}(x)=t_a(-a+x)=a+(-a)+x=x$$ $$t_{-a}\circ t_{a}(x)=t_{-a}(a+x)=-a+(a)+x=x$$

Answer 2

We typically reserve the parenthesis-free notation for linear maps and $t_a$ isn't a linear map. So you should write $t_a(v)$. Also $t_a(v) = a + v$ by definition. We have $t_a(-a) = a - a = 0$ but that isn't important because the rule that a function is one-to-one if and only if its nullspace is $\{0\}$ only applies to linear functions.

Think about this more basically: $t_a$ is the function that adds $a$ so to undo that you subtract $a$.