Uniqueness of inverse of linear maps

Question

Let $V,W$ be real vector spaces, and consider an $\Bbb R$-linear transformation $\varphi:V\to W$.

We know that if $\varphi$ is both left and right invertible, then it admits an inverse and it is unique.

If $\varphi$ is invertible on one side (for example on the right), it can be proved this right inverse is not in general unique.

My question is: if we know that this right inverse is unique, does it imply that $\varphi$ is also left invertible?

I suspect linear spaces/maps don't play any relevant role here; maybe this result, can be discussed for generic functions $f:X\to Y$.

Answer 1

This question provides an answer for $V = W$:

Let $A, B : V \to V$ be linear maps such that $B$ is the unique right inverse for $A$.

We will show that $B$ is also a left inverse for $A$. Consider $BA+B−I$:

$$A(BA+B-I)=ABA+AB-A = A+I-A=I=AB$$

$BA+B-I$ is also a right inverse for $A$. Hence, $BA+B-I=B$, which implies $BA = I$.

Therefore, $A$ is invertible.

Answer 2

I'll show that for a right-inverse to be unique, the kernel of $\phi$ must vanish, and $\phi$ must therefore be injective.

Write $V = V/\ker(\phi)\oplus \ker(\phi)$ and let $\psi : W \to V/\ker(\phi)\oplus\ker(\phi) $ be a right-inverse. Then $\psi$ can be written as $\psi = \psi_1 + \psi_0$, where

• $\psi_1 : W \to V/\ker(\phi)$ and
• $\psi_0 : W \to \ker(\phi)$

Composing $\psi_0$ with any automorphism $\mu : \ker(\psi) \to \ker(\psi)$ yields a right-inverse $\psi_\mu = \psi_1 + \mu \circ \psi_0$ with $\phi \circ \psi_\mu = \phi\circ(\psi_1 + \mu \circ \psi_0) = \phi\circ\psi_1 = \phi\circ(\psi_1 + \psi_0) = id_W$.