Why the empty set isn't a terminal object in the $\mathcal{SET}$ as well as it is an initial object?

Question

It is said that the $\{\}$ is the initial object of the $\mathcal{SET}$ - category of sets and functions.

By definition, it implies that for all $S \in Obj(\mathcal{SET})$ there is exactly one function $f: \{\} \mapsto S$. And that seems more or less all right to me - indeed, there is only one way you can define such $f = \{\} \times S = \{\}$.

The thing I can't get is why $\{\}$ isn't a terminal object? By the same logic, there is only one way to construct $g: S \mapsto \{\} = S \times \{\} = \{\}$.

What am I missing?

P.S. I know that any singleton set is a terminal object in the $\mathcal{SET}$. The question is not about it.

Answer 1

The point is that the subset $f$ of the cartesian product $A\times B$ (for an application $f:A\to B$) must check a property beginning with $$\forall a\in A,\exists b\in B \,\mathrm{s.t.}\dots,$$ so if $A\neq\varnothing$ and $B=\varnothing$, you can't construct a such subset since for a $a$ in $A$, we would have a $b\in\varnothing$, which is impossible.

Answer 2

The empty "function" $g$ you've constructed isn't a function $S \rightarrow \emptyset$ unless $S$ is empty. If $S$ is not empty then there is an element $x \in S$ which doesn't have an output so $g$ is not a function.

Therefore there are no functions from $S$ into the empty set, rather than a unique function.

Answer 3

If $A\neq\varnothing$ then no function $A\to\varnothing$ exists.

This is because for $a\in A$ there is no $f(a)\in\varnothing$.

So $\varnothing$ is not terminal in the category of sets.