Is this thing really a category?

i am having difficulty understanding why the diagram on page 8 of this presentation is a category. the author claims that it is on page 43.

it looks like the two smaller arrows on the left must be idempotent or we would have an infinite number of arrows there. it would also seem that composing one of the smaller arrows on the left with one of the longer (left pointing) arrows must result in the small arrow. composing either left with right must be the identity on the set on the right

i have some code and tests run fine with the categories with 0,1,2,3, and 4 elements as well as another simple example. in these examples the objects are just things.

one of the users here (mike stay) came up with a concrete example using 2 sets for the things. the thing on the left is a set with two elements where each of the small arrows pick out one element. and the thing on the right is a set with a single element. where the long right arrow maps everything to that single element and the two long left arrows map to one of the elements of the set on the right.

A={a1,a2}
B={b1}
short1(x)=a1, short2(x)=a2
right(x)=b1
left1(x)=a1
left2(x)=a2.

i define short1 o left1 = short1 etc. and right(left?) = identity.

but this causes my tests for composition to fail since the source(short1 o left1)!=source(short1).

associativity seems to fails also in a manner similar to the the thing on the right here

the tests fails if i omit left2 from the category.

edit - @MartianInvader 's idea seems to work. the composition table for:

r:A->B, s1:A->A, s2:A->A, l1:B->A, l2:B->A

is

iA·iA=iA
iA·s1=s1
iA·s2=s2
iA·l1=l1
iA·l2=l2
iB·iB=iB
iB·r=r
r·iA=r
r·s1=r
r·s2=r
r·l1=iB
r·l2=iB
s1·iA=s1
s1·s1=s1
s1·s2=s1
s1·l1=l1
s1·l2=l1
s2·iA=s2
s2·s1=s2
s2·s2=s2
s2·l1=l2
s2·l2=l2
l1·iB=l1
l1·r=s1
l2·iB=l2
l2·r=s2

edit: seems like there are two ways to make this thing a category:

one way:

    iA  iB  r   s1  s2  l1  l2  

iA  iA  ·   ·   s1  s2  l1  l2  
iB  ·   iB  r   ·   ·   ·   ·   
r   r   ·   ·   r   r   iB  iB  
s1  s1  ·   ·   s1  s1  l1  l1  
s2  s2  ·   ·   s2  s2  l2  l2  
l1  ·   l1  s1  ·   ·   ·   ·   
l2  ·   l2  s2  ·   ·   ·   ·

another way:

    iA  iB  r   s1  s2  l1  l2  

iA  iA  ·   ·   s1  s2  l1  l2  
iB  ·   iB  r   ·   ·   ·   ·   
r   r   ·   ·   r   r   iB  iB  
s1  s1  ·   ·   s1  s1  l2  l2  
s2  s2  ·   ·   s2  s2  l1  l1  
l1  ·   l1  s2  ·   ·   ·   ·   
l2  ·   l2  s1  ·   ·   ·   ·   

2 Answers

  1. Laurent- Reply

    2019-11-14

    Since you wrote "short1 $\circ$ left1", you must be using $\circ$ to denote composition in the usual order for functions, where $f\circ g$" means first apply $g$ and then apply $f$. But in your "tests for composition", you seem to be using $\circ$ to mean composition in the opposite order, following the order of the arrows instead. I conjecture that, if you used one convention consistently, your problem(s) would disappear.

  2. Lawrence- Reply

    2019-11-14

    There is a problem when you say "it would also seem that composing one of the smaller arrows on the left with one of the longer (left pointing) arrows must result in the small arrow."

    The category you've given is ok, but you are confused about the compositions. Given morphisms $f : A \to B$ and $g : B \to C$, the composition $g \circ f$ is a morphism $A \to C$. That is, the source of $g \circ f$ is the same as the source of $f$. If you apply this to $\text{short1} \circ \text{left1}$ you will see that all is well.

Leave a Reply

Your email address will not be published. Required fields are marked *

You can use these HTML tags and attributes <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>