The function from the $\{\}$ to an any other set?

Question

In the $\mathcal{SET}$ - category of sets and maps between them - there is an initial object - the $\{\}$. It means that there is unique map from the $\{\}$ to an any other set (object of $\mathcal{SET}$).

Not sure I understand it completely. A map $\{\} \mapsto S \in Obj(\mathcal{SET})$ must take some $e \in \{\}$ to a some other $s \in S$. However, there is no such $e$ by the definition of the empty set. Thus such function does exist, however why is it unique? As long as particular $S$ contains more that one element, there are many maps of $\{\} \mapsto S$ kind. Thus the only way to prove that all those maps are essentially the same is to compare them to each other. In order to run comparsion, one has to kind of "compute" those functions - i.e. to assign some $e \in \{\}$ to each and ensure that outputs $s_0, s_1, ..., s_n \in S$ are the same. Otherwise, how can you determine that abovementioned functions are equal bypassing their evluation?

Answer 1

Think of it this way: we say two functions $f$ and $g$ on the same domain are different if there exists an $x$ in that domain so that $f(x) \neq g(x)$. In the case where the domain is empty, there obviously cannot be such an $x$, so $f$ and $g$ are automatically not different.

More formally, a function is formally defined as a set of ordered pairs $(e, f(e))$, where $e$ is taken from the domain. If the domain is empty, no such pairs can exist - so any function with domain $\{\}$ must itself be the empty set. And two "copies" of the empty set are automatically equal.

Answer 2

Functions from $A$ to $B$ are subsets of $A\times B$. In particular they are sets. The only function from $\emptyset$ to any other set $S$ is the empty function $\emptyset\subset\emptyset\times S$.

Warning: opinionated! (read comments below)

More strikingly, there isn't just one function for every $S$, it is also the same for all $S$!

Answer 3

A map {}↦S∈Obj(SET) must take some e∈{} to a some other s∈A (emphasis mine)

Not quite. A map $X\to Y$ must take every $e \in X$ to some $s \in Y$.

Every does not imply some. some always means more than $1$. But if $X = \emptyset$ then every means $0$ or none. If there are no $e \in X = \emptyset$ to begin with then every element of $\emptyset$ is being mapped to an element of $Y$. That is because there aren't any $e \in X$ so there aren't any $e \in X$ that aren't being mapped. So all of them are being mapped; all $0$ of them.

In the same sense all elements of $\emptyset$ have a first class ticket to Mars on a unicorn made of pink champagne. And all elements of $\emptyset$ are staying at home playing the tuba. All elements of $\emptyset$ are doing anything and everything and not doing anything and everything.

This is called vacuous truth. Everything is true about every $e \in \emptyset$ because there aren't an $e \in \emptyset$ so there aren't any cases where it is false.

Okay.... what does this empty $i: \emptyset \to Y$ look like?

Consider a typical function $f: X\to Y$. Technically $f= \{e\mapsto e_y, d\mapsto d_y,.....\}$ or in other words $f$ is a set of ordered mappings. An ordered mapping $e\mapsto e_y$ can be written as an ordered pair $(e,e_y)$ so a function $f$ is a set of ordered pairs.

So $f:X\to Y$ is a set $f\subset X\times Y = \{(x,y)|x\in X, y\in Y\}$. There are some rules. For every $x \in X$ there is a $y\in Y$ so that $(x,y) \in f$. And if $(x,y)\in f$ then for any $w \ne y$ then $(x,y)\not \in f$. But those rules aren't important right now. The important thing now is $f \subset X\times Y$.

That's what a function "really" is. You can think of it as some kind of rule, or a little box with a handle that when you put in input output comes out, or some little monster with a pointing tail. But what it technically is is a set of ordered pairs. (All these interpretations are compatible.)

So what is $\emptyset\times Y$? Well, it is $\{(x,y)|x\in \emptyset, y\in Y\}$ but as there are no $x \in \emptyset$: $\emptyset\times Y=\{(x,y)|x\in \emptyset, y\in Y\}=\emptyset$

So $i:\emptyset \to Y$ is simply... an emptyset of zero ordered pairs.

One way to think about it is... the empty function just is. It maps nothing to anything you want and you don't have to do anything to have it happen.

In a way, it's useless because nothing can be afffected or mapped by it. But logically from a constructivist point of view we must build everything up from nothing and a foundation for all objects must be an empty object. So we must start with an empty function that... does nothing.