I have been studying category theory for fun, and got confused on a concept. So apparently we can view a group as a one object category where the morphisms are the group elements and we define the composition of maps by the product of the group elements. This all makes sense to me, till I get really technical. For example, a morphism is a function from one object to another. So let us look at $Z_{3}$, we have the group element 1. If we want to think of 1 as a morphism we have to technically define a map from $Z_{3}$ to $Z_{3}$. Since the group is cyclic I will define the map $1:Z_3 \rightarrow Z_3$ by $0\rightarrow 1$, $1\rightarrow 2$, and $2 \rightarrow 0$. Similarly, define the map $2:Z_{3}\rightarrow Z_{3}$ by $0 \rightarrow 2$, $1 \rightarrow 0$, and $2 \rightarrow 1$, and lastly define $0$ as the identity map. The compositions of the maps will act as the group composition of the elements. Anyways, the construction of these maps are easy since $Z_{3}$ is cyclic. But how would I use the same concept for example on the group of reals under addition? Am I being to technical? Am I taking the definition of the morphism to literal?

  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun Feb 7 at 20:24
  • $\begingroup$ Some paragraphing might help . . . $\endgroup$ – Shaun Feb 7 at 20:24

If a group is thought of as a category with just one object, which we might denote *, then an element of the group becomes a morphism from * to itself (so is an automorphism of the object *).

The `category' $\mathbb{Z}_3$ has a single object. The three elements of $\mathbb{Z}_3$ are morphisms from * to *. 0 is the identity morphism on * and $1:*\to *$ is another morphism, and the composition of $1$ with itself gives us $2:*\to *$. NB: * is just an abstract object and is not $\mathbb{Z}_3$ as you tried to write. About the only thing you can know about this object is that it has exactly two endomorphisms other than the identity and they are both invertible, so they are abstract automorphisms. The automorphisms of * form a group isomorphic to $\mathbb{Z}_3$.

I think your final questions are based on a confusion, so they do not quite make sense.

The related question: Confused about the definition of a group as a groupoid with one object. may help.

  • $\begingroup$ Hello Tim, originally I thought of exactly what you wrote down. But this is what confused me. When you wrote "$1:*\rightarrow *$ is another morphism." Don't you have to tell me where you are the sending the elements in $*$ since that set is the domain? $\endgroup$ – GentGjonbalaj Feb 7 at 20:43
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    $\begingroup$ No. The domain of 1 is the object *. That is NOT the set $\mathbb{Z}_3$. It is an abstract object. I will write $\mathcal{C}$ for the category so $Ob(\mathcal{C}) =\{*\}$. We have to specify the sets of morphisms between any pair of objects of this category... there is only one so we need to write down $\mathcal{C}(*,*)$ and that will be $\{0,1,2\}$, i.e. the set $\mathbb{Z}_3$ , then we need to specify the composition of each composable pair of elements and that will be given by the addition in $\mathbb{Z}_3$. I am a bit short of space so will leave the rest, but ask if someone needs more. $\endgroup$ – Tim Porter Feb 7 at 20:50
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    $\begingroup$ It seems like you are just defining three morphisms without a care of what we are mapping to what. We just care that the composition of the maps behave how the group elements would behave. Hopefully I got it now. Thanks for the help Tim. $\endgroup$ – GentGjonbalaj Feb 7 at 20:57
  • $\begingroup$ @GentGjonbalaj Just to elaborate, the morphisms here aren't "mapping" anything. The object $*$ is not a set, and morphisms $* \to *$ are not functions. What are they? Just abstract things that you can compose (and the compositional structure is isomorphic to $\mathbb Z_3$). In many categories the objects are indeed sets, and the morphisms are (certain sorts of) functions, and composition of morphisms is just composition of functions, but this is not true in general. $\endgroup$ – Daniel Mroz Feb 7 at 22:25
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    $\begingroup$ @GentGjonbalaj I would consider that a (mild) abuse of notation, or maybe just unfortunately suggestive. The point of arrows having a domain and codomain is so that you can specify when composition is legal and when it isn't. Arrows in a category behave (in some respects) like functions on sets, but they don't have to be functions on sets, they're just things that can be composed. $\endgroup$ – Daniel Mroz Feb 7 at 22:42

A morphism is not necessarily a function from one object to another, this is merely the generic situation. All that is really required is that they can be composed.

In any case, the maps that you're writing down are not weirdly technical, they are closely related to the notion of a group action. Given whatever group $G$, the category that encodes that group has a single group $G$ and its morphisms consist of all the functions $y \mapsto x \cdot y$ where $\cdot$ is the group operation and $x$ is an arbitrary group element. The composition is then just function composition.


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