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

  • 1
    $\begingroup$ My advice would be to ignore this comment for now. If you understand the definition of a terminal object, you are good to go. You can come back to that comment when you see the notion of limit (which the product is an example of), then you will see the relevance of that observation. $\endgroup$ – Pece Nov 2 '16 at 8:24
  • $\begingroup$ I take your advice. Thank you. I am reading about limits right now so I hope this will make sense in the following couple of hours :) $\endgroup$ – Vinyl_cape_jawa Nov 2 '16 at 9:52

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.

  • $\begingroup$ A most excellent reply! :-) $\endgroup$ – Musa Al-hassy Nov 22 '18 at 10:39

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 $0$.

  • $\begingroup$ so here ! is this "empty" function? $\endgroup$ – Vinyl_cape_jawa Oct 30 '16 at 13:52
  • $\begingroup$ I cannot recognize this as an answer to the question, that deals with nullary products (not nullary functions). $\endgroup$ – drhab Oct 30 '16 at 14:10
  • $\begingroup$ @drhab: And an $n$-ary product can be expressed as a function (functor even!) on $n$ inputs. But really, the only thing the OP needs is to see the idea of a calculation with zero inputs; it doesn't really matter what is being calculated. $\endgroup$ – Excluded and Offended Nov 1 '16 at 20:35
  • $\begingroup$ I see now and took away my downvote. This was only possible after an edit, so I took the liberty to add some info. I hope you don't mind. $\endgroup$ – drhab Nov 2 '16 at 8:13
  • $\begingroup$ To clarify, morphisms $1 \to X$ are known as points, and in set there are as many of them as the number of elements in $X$. This is related to but different from what OP asks. $\endgroup$ – Musa Al-hassy Nov 22 '18 at 10:41

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".


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.


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