Your confusion seems to be not so much about initial and terminal objects, but about what those look like in the category of sets. Looking at the formal definition of “function” will help make clear some of the unusual cases such as functions with empty domains.
A function from $A$ to $B$ can be understood as a set of pairs $$\langle a,b\rangle$$ where $a\in A$ and $b\in B$. And:
There must be exactly one pair $\langle a,b\rangle$ for each element $a$ of $A$.
Exactly one, no more and no less, or the set of pairs is not a function.
For example, the function that takes an integer $n$ and yields its square $n^2$ can be understood as the (infinite) set of ordered pairs:
$$\{ \ldots ,\langle -2, 4\rangle,
\langle -1, 1\rangle,
\langle 0, 0\rangle ,\langle 1, 1\rangle,
\langle 2, 4\rangle\ldots\}$$
And for each integer $n$ there is exactly one pair $\langle n, n^2\rangle$. Some numbers can be missing on the right side (for example, there is no pair $\langle n, 3\rangle$) and some numbers can be repeated on the right (for example the function contains both $\langle -2, 4\rangle$
and $\langle 2, 4\rangle$) but on the left each number appears exactly once.
Now suppose $A$ is some set $\{a_1, a_2, \ldots\}$ and $B$ is a set with only one element $\{b\}$. What does a function from $A$ to $B$ look like? There is only one possible function: it must be: $$\{
\langle a_1, b\rangle,
\langle a_2, b\rangle,
\ldots\}.$$ There is no choice about the left-side elements of the pairs, because there must be exactly one pair for each element of $A$. There is also no choice about the right-side element of each pair. $B$ has only one element, $b$, so the right-side element of each pair must be $b$.
So, if $B$ is a one-element set, there is exactly one function from $A$ to $B$. This is the definition of “terminal”, and one-element sets are terminal.
Now what if it's $A$ that has only one element? We have $A=\{a\}$ and $B=\{b_1, b_2, \ldots\}$. How many functions are there now? Only one?
One function is $$\{\langle a, b_1\rangle\}$$
another is $$\{\langle a, b_2\rangle\}$$
and another is $$\{\langle a, b_3\rangle\}$$
and so on. Each function is a set of pairs where the left-side elements come from $A$, and each element of $A$ is in exactly one pair. $A$ has only one element, so there can only be one pair in each function. Still, the functions are all different.
You said:
I would find it more intuitive if one-element set would be initial object too.
But for a one-element set $A$ to be initial, there must be exactly one function $A\to B$ for each $B$. And we see above that usually there are many functions $A\to B$.
Now we do functions on the empty set. Suppose $A$ is $\{a_1, a_2, \ldots\}$ and $B$ is empty. What does a function from $A\to B$ look like? It must be a set of pairs, it must have exactly one pair for each element of $a$, and the right-side of each pair must be an element of $B$. But $B$ has no elements, so this is impossible: $$\{\langle a_1, ?\rangle, \langle a_2, ?\rangle, \ldots\}.$$
There is nothing to put on the right side of the pairs. So there are no functions $A\to\varnothing$. (There is one exception to this claim, which we will see in a minute.)
What if $A$ is empty and $B$ is not, say $\{b_1, b_2, \ldots\}$? A function $A\to B$ is a set of pairs that has exactly one pair for each element of $A$. But $A$ has no elements. No problem, the function has no pairs! $$\{\}$$
A function is a set of pairs, and the set can be empty. This is called the “empty function”. When $A$ is the empty set, there is exactly one function from $A\to B$, the empty function, no matter what $B$ is. This is the definition of “initial”, so the empty set is initial.
Does the empty set have an identity morphism? It does; the empty function $\{ \}$ is its identity morphism. This is the one exception to the claim that there are no functions from $A\to\varnothing$: if $A$ is also empty, the empty function is such a function, the only one.
The issue for topological spaces is exactly the same:
- When $B$ has only one element, there is exactly one continuous map $A\to B$ for every $A$.
- When $A$ is empty, there is exactly one continuous map $A\to B$ for every $B$: the empty function is the unique map.
- When $A$ has only one element, there are usually many continuous maps $A\to B$, one different one for each element of $B$.
There are categories in which the initial and terminal objects are the same. For example, in the category of groups the trivial group (with one element) is both initial and terminal.
I hope this was some help.
Best Answer
Your problem is that your intuitions about the empty set don't fit the mathematical definitions; where functions etc. end up existing vacuously. In this answer, given your background I'll discuss the category Set. This category has as its objects sets and as its morphisms functions between sets.
In this setting, your question asks what the morphisms with domain $\emptyset$ are. The point is that the definition of a function $f:A \to B$ is "for each $a \in A$ there is a unique $b \in B$ such that $f(a) = b$" (In set theoretic language $f:A \to B$ is actually a subset of $A \times B$ such that for each $a \in A$, there is a unique $b \in B$ such that $(a,b) \in f$).
In particular, this definition means that for any set $B$ there is a unique function from $\emptyset$ to $B$; namely $\emptyset$ itself! (though in the category we distinguish between the object $\emptyset$ and each of the morphisms $\emptyset$ since the morphisms come with a domain and codomain) This may seem strange, but the definition requires it since there are no $a \in \emptyset$ so the needed condition vacuously holds for $f = \emptyset$. In particular, in Set for every object $B$ there is a unique morphism $f:\emptyset \to B$ (In technical language, this says that $\emptyset$ is an initial object in Set). So, yes, in Set, $\emptyset$ has an identity morphism $\emptyset \to \emptyset$ and also participates in morphisms to all other sets. This means your intuition about the category Set is incorrect.
In general, a category is really just a collection of objects with arrows (morphisms) satisfying the right rules. Anything you write down with those properties is fine! We could consider the category whose objects are sets with just the identity arrows, or with all possible arrows; both would be fine. The advantage of this is that it lets lots of things fit into this framework.