If $T: \mathcal{V}\to\mathcal{W}$ is a transformation, and $T^{-1}$ exists, then $T$ is linear

Question

Is this always true? Is there an example of a non-linear transformation such that its inverse is well-defined? Thanks.

Answer 1

What about $T:{\bf{R}}\rightarrow{\bf{R}}$, $T:x\rightarrow x+1$, then $T^{-1}:x\rightarrow x-1$.

Answer 2

No, absolutely not.

If $T^{-1}$ exists, it just means that $T$ is a bijection, which means that for every $y\in \mathcal W$, there is a unique $x\in \mathcal V$ such that

$$T(x)=y.$$

Or if you prefer, it is equivalent to the fact that there exist

$$U\colon \mathcal W\to\mathcal V$$

such that

$$T\circ U=\mathrm{id} \text{ and } U\circ T=\mathrm{id}.$$

If you want a counterexample, take

$$x\mapsto x^3$$

from $\mathbb R$ to $\mathbb R$, or

$$x\mapsto e^x$$

from $\mathbb R$ to $\mathbb R_+^*$.