Is the empty function its own inverse

Question

I suspect that the empty function $f: \emptyset \to \emptyset$ is self-invertible, i.e., $f^2 = \mathrm{id}$, but I don't know how I would go about proving it. I would need to show that for every $x \in \emptyset$, $(f \circ f) (x) = x$. But there are no such $x$. If the condition for an inverse is an implication (if $x$ is an element of the empty set, then this property holds), the condition is vacuously true, but is that enough? Alternatively, a function is its graph, and $f$ is just an empty list. The inverse of the empty list seems that it should be the empty list again, but I don't know precisely how to "prove" that intuition.

Answer 1

Since $\emptyset$ contains no element, the assertion $(\forall x\in\emptyset):(f\circ f)(x)=x$ is vacuously true. So, the empty function is indeed its own inverse.