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So I generally understand the idea of a monoid from set theory, but I'm trying to understand better the mapping to category theory.

http://en.wikipedia.org/wiki/Monoid#Relation_to_category_theory

I think there are a couple of things that are tripping me up, and I think it'd be instructive to have a concrete example, so let's do the monoid of integers. So the associative operator is +, and the zero is 0.

As a category, it has one object.. what is that object? And according to the wikipedia link, we have morphisms for every element... so we have a category whose object is Int, and the morphisms are 0, 1, 2, 3... (important question: can an object have multiple distinct morphisms? I didn't think that was defined, as I thought that morphisms were completely defined by the domain and codomain). So now we need our associative operator, ideally such that the map named 1 composed with the map named two is the same as the map named 3. I think this is where there is a breakdown in how I am thinking about these. How do we think about that last property?

Thank you very much for any help you have.

Update: the answer given gets me 95% of the way there, but there is one lingering question: The one thing I'm not sure about is how do we formalize equivalence? Like, we can definitely say that (a * b) * c = a * (b * c) (by the definition of a category), but how do we "define" 1 * 2 = 3 and think about it in a rigorous way when they are all morphisms with the same domain and codomain. Another way of asking this is if we have a morphism called 2, and a morphism called 3, and a morphism called 7, how do we know that 2*3=7?

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  • $\begingroup$ I'm not sure I understand your update, but would it help to think of the following example? A category with one object $S$, which is a set, and the morphisms are all functions from $S$ to $S$. The "composition" operation is composition of functions. $\endgroup$ – user180040 Oct 4 '14 at 14:36
  • $\begingroup$ My question is this: how do we know that 1+2=3 but 1+2 != 7? How do we reason about the values of the compositions of these morphisms? $\endgroup$ – A Question Asker Oct 4 '14 at 17:04
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    $\begingroup$ When defining a category, you get to choose what the composition operation is for morphisms. $\endgroup$ – user180040 Oct 4 '14 at 17:07
  • $\begingroup$ Really? I thought all morphisms had to compose! $\endgroup$ – A Question Asker Oct 4 '14 at 17:46
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    $\begingroup$ Considering your update: Define a category $\mathcal{N}$ via: $\operatorname{Ob}(\mathcal{N}) := \{*\}$, $\operatorname{Hom}_\mathcal{N}(*,*) := \{f_i|i\in\mathbb{N}\}$, and composition $f_i\circ f_j = f_{i+j}$. This might seem cheap, but it's easy to see that this $\mathcal{N}$ satisfies every axiom for a category (of course this is only one representant for the categorical monoid, since objects in $\mathcal{Cat}$ are only known up to isomorphisms). $\endgroup$ – roman Oct 7 '14 at 8:20
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It doesn't make a difference what the object is, so I doubt anybody has ever specified what it is. This is the same kind of thing as when you consider a "one-point space" $\{*\}$ in topology.

You said that a morphism is determined by its domain and codomain. (Or source and target object, so to speak.) That's not true, even in the category of sets. There's usually more than one mapping from a set $A$ to a set $B$.

For the last thing, what you wrote is correct. Of course $1$ and $2$ are not true "mappings," but they're what stands in for mappings in this category. If $1$ and $2$ are thought of as analogous to mappings, then the operation $+$ should be thought of as analogous to $\circ$. So $1 \circ 2 = 3$, etc.

By the way, this category has a concrete realization as the category with a single object $\mathbf{Z}$ whose morphisms are all automorphisms of $\mathbf{Z}$ as an ordered set.

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    $\begingroup$ Thank you! The one thing I'm not sure about is how do we formalize equivalence? Like, we can definitely say that (a * b) * c = a * (b * c) (by the definition of a category), but how do we "define" 1 * 2 = 3 and think about it in a rigorous way when they are all morphisms with the same domain and codomain. $\endgroup$ – A Question Asker Oct 4 '14 at 14:33
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You already know what the integers with addition are.

When you construct this category, you define composition to be addition. There's not a problem doing that, because you already know what addition is.

important question: can an object have multiple distinct morphisms

Yes, if I understand you: in general, the set of homomorphisms from $X$ to $Y$ may have many elements. One of the typical examples of a category is $\mathbf{Set}$, the category whose objects are sets and the whose morphisms are functions.

I this category, the domain and codomain of a function in the sense of the category is its given by its domain and codomain in the sense of set theory. (I'm assuming a definition of 'function' that includes the domain and codomain as part of the definition. If you pick a different definition: e.g. that a function is simply the graph of a relation, you'll have to modify the definition of morphism in $\mathbf{Set}$ appropriately)

And in $\mathbf{Set}$, the product of two morphisms is their composite.

In this category, we can easily see that $\hom(X,Y)$ generally has lots and lots of elements, except in special cases. (e.g. $X = \varnothing$ or $|Y| = 1$)

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