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
 A: 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 $0$.
A: 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".
A: 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.
A: In general, a product of a family of objects $A_i$ is an object $\prod_iA_i$ together with maps $\pi_i:\prod_iA_i\to A_i$ such that any other such object $B$ with maps $f_i:B\to A_i$, there exists a unique map $\phi:B\to \prod_iA_i$ such that $f_i=\pi_i\phi $ for all $i$. Now what happens with an empty family? Well we only require to have an object $1$, without any special map, and for any other object without any special map, there must be a unique map $B\to 1$, but no special condition on this map. So $1$ is precisely a terminal object.
