Left inverse property implies two-sided inverses exist

Statement
Suppose $$(L,*)$$ is a loop with neutral element $$e$$. Suppose $$L$$ is a left inverse property loop, i.e., there is a bijection $$\lambda:L \to L$$ such that for every $$a,b \in L$$, we have:

$$\! \lambda(a) * (a * b) = b$$

Then, $$\lambda(x)$$ is the unique two-sided inverse of $$x$$ (in a weak sense) for all $$x \in L$$:

$$\! \lambda(x) * x = x * \lambda(x) = e$$

Note that it is not necessary that the loop be a right-inverse property loop, so it is not necessary that $$\lambda(x)$$ be a right inverse for $$x$$ in the strong sense.

Related facts

 * Equality of left and right inverses in monoid
 * Two-sided inverse is unique if it exists in monoid
 * Equivalence of definitions of inverse property loop
 * Equivalence of definitions of gyrogroup

Proof
Given: A left-inverse property loop $$L$$ with left inverse map $$\lambda$$.

To prove: $$\! \lambda(x) * x = x * \lambda(x) = e$$, where $$e$$ is the neutral element.

Proof: Putting $$a = x, b = e$$ in the left inverse property condition, we obtain that $$\lambda(x) * x = e$$.

Next, putting $$a = x, b = \lambda(x)$$, we obtain that:

$$\! \lambda(x) * (x * \lambda(x)) = \lambda(x)$$

Writing the $$\lambda(x)$$ on the right as $$\lambda(x) * e$$ and using cancellation, we obtain that:

$$\! x * \lambda(x) = e$$

This completes the proof.