Is the empty set homeomorphic to itself?

Solution 1:

Your map $h$ does exist, and is a homeomorphism. In fact, it's the identity map: for every element $x\in\emptyset$, $h(x)=x$. So since $h\circ h=h=\operatorname{id}$, $h$ is its own inverse. Since both $h$ and $h^{-1}=h$ are continuous, $h$ is a homeomorphism.

(Incidentally, checking that $h$ is continuous isn't entirely vacuous. You have to check that $h^{-1}(U)$ is open for any open subset $U\subseteq\emptyset$. It is not true that there are no choices of $U$: rather, there is exactly one choice of $U$, namely $U=\emptyset$. Of course, $h^{-1}(\emptyset)=\emptyset$ is indeed open.)

Solution 2:

fleshing it out a little more, there is the empty function $\emptyset \to \emptyset$ ( that is just $\emptyset$ viewed as a function). This is bijective and bicontinuous.

Solution 3:

In every category, every object is isomorphic to itself, via the identity morphism. This follows immediately from the definitions of category theory, the existence of a identity morphism for every object is an axiom. So if you want to call topological spaces a category, and the empty space an object of that category, then you must accept that the identity morphism of that space is a homeomorphism. The same goes for the empty set as object in the category of sets, whose identity map is a bona fide bijection.

By the way "identity" morphism (or arrow) is just a name, and does not have to be an identity map (or a map at all), though it does have to be neutral in the composition with other morphisms (and with itself). But in the categories of sets and topological spaces morphisms are maps, and the identity map is the only map that is neutral in composition with morphisms, so the identity morphism must be the identity map. And in case of an empty set/space, the identity map is the only map to itself around anyway.