Re: Are there exactly 11 categories with 3 arrows?

[David Roberts <droberts@maths.adelaide.edu.au>]

Hi Mark,

Let us phrase the question as:

Given three sets, {a},{b,c},{d,e,f}, how many categories are there
with one of these sets as its set of objects and exactly three arrows?
(any other set of objects is disallowed, as you pointed out)

This is to remove questions of 'up to equivalence' because
technically, there are an infinite number of categories with three
arrows, and taking categories up to equivalence will, I think, give
too few for your liking (for example, in my list of four categories
with two objects, we would only get two categories).

Here's a start: there are two categories b -> c, c -> b (omitting
identity arrows) with no non-trivial isomorphisms, and two categories
Z/2 \sqcup {*} where * is one of b or c, and the other is the identity
element of Z/2. Given your unique category with three elements this
brings us to five categories. Then there is Z/3 (with identity element
a), giving us six.

What other categories with three arrows have you identified? They
will, I imagine, have two objects.

David



On 27 March 2010 00:37, Mark Spezzano <mark.spezzano@chariot.net.au> wrote:
> Hi,
>
> This is a beginner's question. I have a textbook (Walters) that asks to show that there are exactly 11 categories with 3 arrows.
>
> Now, my logic tells me that I need to cover three possibilities:
>
> a) One object with three arrows. How many are there of these?
>
> b) Two objects with three arrows. How many are there of these?
>
> c) Three objects with three arrows. I think that the answer to this is the easiest. The answer is 1 categories because they are all endomorphisms, each object containing just the identity morphism.
>
> So the other 10 arrows must come from a) and b), but I keep getting different answers like 12 and 13 categories as the total.
>
> Can someone please explain to me the combinations of categories that need to be covered and why some are missed out during the calculation.
>
> Any help would be immensely appreciated.
>
> Thanks,
>
> Mark Spezzano
>
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