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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?

user404735
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  • 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. – Shaun Feb 07 '19 at 20:24
  • Some paragraphing might help . . . – Shaun Feb 07 '19 at 20:24

2 Answers2

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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.

Tim Porter
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  • 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? – user404735 Feb 07 '19 at 20:43
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    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. – Tim Porter Feb 07 '19 at 20:50
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    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. – user404735 Feb 07 '19 at 20:57
  • @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. – Daniel Mroz Feb 07 '19 at 22:25
  • @DanielMroz I thought a morphism must be a mapping because in the book I am using, the definition of a category says, " for each arrow $f$ there are given objects dom(f) and cod(f)." So if the arrows have domains and codomains isn't it clear that the arrows are technically always a mapping? – user404735 Feb 07 '19 at 22:29
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    @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. – Daniel Mroz Feb 07 '19 at 22:42
  • @gent it seems that your hang up is the fact that ‘morphisms’ don’t actually need to be ‘functions’ in the set theoretic sense. This is one of the abstractions if category theory. For example any poset can be viewed as a category where you have a morphism A to B iff A leq B – Prince M Feb 08 '19 at 03:22
  • Oops, I just realized I repeated Daniels last comment. It was buried when I posted. – Prince M Feb 08 '19 at 03:23
  • @PrinceM But wait if we have a morphism A to B iff A leq, you described the mapping. Every example I have ran into a mapping is described. It is really weird to talk about these abstract objects morphisms where we describe no mapping of elements. It sounds like we are talking about monoid then. – user404735 Feb 08 '19 at 03:54
  • But I didn’t describe the mapping because A and B aren’t even sets, they don’t even have elements. I’m just saying “if A is leq B in your poset, draw an arrow from A to B in your category”. The morphisms of category’s are more like the edges of graph theory, although it just so happens that a lot of times they are functions – Prince M Feb 08 '19 at 04:46
  • @GentGjonbalaj: A monoid `is', similarly, a category having a single object and given any category $C$ and any object $x$ in $C$, the set $C(x,x)$ of arrows from $x$ to itself forms a monoid under the composition coming from the category. – Tim Porter Feb 08 '19 at 07:46
  • It may help to return to the poset example in a bit more detail and perhaps a bit more formally. Let $(P,\leq)$ be a poset. Define a category $\mathbb{P}$,using this data. An object of $\mathbb{P}$ is an element of the set $P$. Let $a$ and $b$ be two elements of $P$, hence objects of $\mathbb{P}$.There is to be an arrow $a\to b$ in $\mathbb{P}$ if, and only if, $a\leq b$ . We still have to define composition, but if $a\to b$ and $b\to c$ in $\mathbb{P}$, then $a\leq b$ and $b\leq c$ in $(P,\leq)$, but by transitivity of $\leq$ we know this means $a\leq c$ so there is a unique arrow $a\to c$. – Tim Porter Feb 13 '19 at 16:47
  • Identities simply correspond to $a\leq a$, which we know is true and associativity should be easy to check. NB. This applies to any poset so the elements of $P$ need not be sets or anything like them! – Tim Porter Feb 13 '19 at 16:49
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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.

Ian
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