I am quite familiar with preimages, however, I have recently been taking a course on Set Theory and came across questions that involve preimages of functions with empty domains (and sometimes empty codomains too).
I was wondering how we treat these particular cases and whether or not there is a particular convention when it comes to dealing with these particular problems.
To make this more concrete, here is one of the questions that I have come across that prompted me to post my question:
For the (unique) function $f: \emptyset \rightarrow$ {$\emptyset$}, explicitely find the preimage $f^{-1} :${$\emptyset$, {$\emptyset$}} $\rightarrow$ {$\emptyset$}
My thoughts here were that {$\emptyset$} $\mapsto \emptyset$ since there are no elements in the domain of $f$ and so this means that there are no elements that can possibly map to {$\emptyset$}.
I also believe that $\emptyset \mapsto \emptyset$. I am less confident about this. I am looking for the elements of the domain of $f$ that map to nothing. On the one hand, the domain is empty so there is nothing that can possibly map to anything. However, since we are asking what elements of the domain map to nothing, it also feels like the entire domain maps to nothing which makes me think it might instead map the empty set to {$\emptyset$}. Although perhaps I am overthinking it here.
I would be grateful for any clarification.
Best Answer
There is a unique function from the empty set to any set $S$, empty or not - namely, the empty function. For every $s \in S$ the preimage for the empty function is the empty set.
There is some ambiguity in the definition of $f^{-1}$. If that is a function whose domain is the codomain $S$ of $f$ then if $S$ is empty too there are no preimages of elements and $f^{-1}$ is the empty function.
If you think of $f^{-1}$ as a function whose domain is the set of subsets of the codomain of $f$ then when $S$ is the empty set the only subset of $S$ is $\emptyset$ and $$ f^{-1}(\text{$\emptyset$}) = \emptyset . $$
In the question, $S =$ {$\emptyset$} , which is a set with one element. The inverse image of that single element is the empty set since nothing in the (empty) domain maps to it: $$ f^{-1}(\text{$\emptyset$}) = \emptyset . $$
In the second interpretation of $f^{-1}$, $S$ has two subsets and for both
$$ f^{-1}(\text{{$\emptyset$}}) = f^{-1}(\text{$\emptyset$}) = \emptyset . $$