Terminal objects as "nullary" products

I read something weird in my category theory book (Awodey p 47).

" Observe also that a terminal object is a nullary product, that is, a product of no objects:

Given no objects, there is an object $1$ with no maps, and given any other object $X$ and no maps, there is a unique arrow:

$$!:X\to 1$$

making nothing further commute."

Could anyone give a hint about what this means? I mean "given no objects, there is an object..?"

Thank you

To form a product, you give me $n$ objects, $A_1,\dots,A_n$, and I give you back an object $A_1\times\dots\times A_n$, together with $n$ maps $\pi_i\colon A_1\times\dots\times A_n\to A_i$ (one to each of the $A_i$) satisfying the universal property of the product.

So what happens if $n=0$? Then you give me $0$ objects, and I give you back an object which we call $1$, together with $0$ maps $\pi_i$ (one to each of the $A_i$, of which there aren't any), satisfying the universal property of the product.

What does the universal property say in this case?

For any $X$ given together with $0$ maps $f_i$ (one to each of the $A_i$, of which there aren't any), there is a unique map $!\colon X\to 1$ making all of the triangles commute ($\pi_i\circ ! = f_i$ for all $i$, of which there aren't any).

Removing the vacuous conditions from the definition, we see that the empty product is an object $1$ such that for every object $X$ there is a unique map $!\colon X\to 1$, i.e. $1$ is a terminal object.

You are familiar with functions of two variables, such as

$$ f(x, y) = x^2 + y^2 $$

You are familiar with functions of one variable, such as

$$ g(x) = \sin(2x) $$

There are also functions of zero variables, such as

$$ h() = \pi $$

We often call such things "constants" and handle them in a special way — but for various purposes it is convenient to treat them uniformly with other sorts of functions.

A function that takes no arguments is also called a nullary function or a function with arity $0$.

Well, what is a normal product (say of two objects, $A$ and $B$)? It is an object $A\times B$ with maps $\pi_1:A\times B\to A$, $\pi_2:A\times B\to B$ such that given $C$ with maps $f:C\to A$ and $g:C\to B$, there is a unique arrow $h:C\to A\times B$ making whatever diagram commute.

Well, look at our situation with the terminal object $1$. If you (try to) think of it as a product, you will see there is no projection map included (i.e., it is a "product" of nothing). Now, look at the situation in the paragraph above. We need to have an object $C$ with arrows to the objects we took a "product" over. Well, we didn't take a product over any objects, so there does not need to be any arrows, just an object $C$. Then, because $1$ is terminal there is a unique map $C\to1$, which makes the "diagram" commute (there really isn't a diagram though, just the map $C\to 1$).

Hence, $1$ satisfies the requirements of being a "product" only with no actual objects, so it is an "empty product".