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.

short1(x)=a1, short2(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



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  ·   ·   ·   ·   

migrated from mathoverflow.net Jan 31 '14 at 19:35

This question came from our site for professional mathematicians.

  • 1
    $\begingroup$ Hi rtayek: For writing on MathOverflow, can I recommend that you find the shift key on your keyboard? There are many "good" styles of writing on the internet, but MathOverflow tends to work best in a somewhat "professional" (academic) style, with standard capitalization and so on. That said, the content of your question is probably better suited for Math.StackExchange (where I believe the aesthetics also favor standard academic English over "internet English"). $\endgroup$ – Theo Johnson-Freyd Jan 31 '14 at 19:10

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.

  • $\begingroup$ sorry, i can not see the inconsistency. tests for composition says: short1 o left1. applying left1 first, this is equal to short1 by definition. but the source of short1 o left1 should be the source of left1. but they are different. $\endgroup$ – Ray Tayek Jan 31 '14 at 19:44
  • $\begingroup$ No they are not. $\endgroup$ – Andrej Bauer Jan 31 '14 at 20:04

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.

  • $\begingroup$ yes, source(short1 o left1) = source(left1). but if i define short1 o left1 = short1, the sources are not equal. i can't seem to see any other way to define short1 o left1. if it's not equal to s1 (or s2), i am missing an arrow and it's not a category :( $\endgroup$ – Ray Tayek Jan 31 '14 at 21:34
  • $\begingroup$ It seems the error is yours, you're defining short1 $\circ$ left1 $=$ short1, but it seems that it should be left2. $\endgroup$ – MartianInvader Jan 31 '14 at 21:59
  • $\begingroup$ i did not consider that. i will try it and see what my code does. $\endgroup$ – Ray Tayek Feb 1 '14 at 2:05
  • $\begingroup$ that seems to work. see edit $\endgroup$ – Ray Tayek Feb 1 '14 at 2:39
  • $\begingroup$ Why are you even defining composition? Just use composition of functions and calculate what you get. $\endgroup$ – Andrej Bauer Feb 1 '14 at 7:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.