Let $\mathbb I$ be a category with exactly $2$ objects and $4$ arrows. The $2$ arrows that are not identities are parallel and their (common) domain and (common) codomain are distinct. Looking at the category $$\mathsf Set^{\mathbb I}$$ (having as objects the functors $\mathbb I\rightarrow\mathsf Set$ and as arrows the natural transformations) I wondered if this category can be recognized as the category of (small) graphs. Is that indeed the case, or am I overlooking something?

If I am right here then the fact that $\mathsf Set$ is (co)complete justifies the the conclusion that $\mathsf Grph$ is (co)complete.

  • $\begingroup$ A functor $\Bbb I\to\mathbf{Set}$ is then determined by the choice of two isomorphic sets and one bijection between them. $\mathbf{Set}^{\Bbb I}$ is just the category of bijections, whose arrows are the set maps which are compatible with these bijections. $\endgroup$ – Stefan Hamcke Mar 20 '14 at 14:25
  • $\begingroup$ @StefanHamcke That would be so if the non-identities are inverses of eachother right? But they are not. They are parallel. $\endgroup$ – H.B. Mar 20 '14 at 14:29
  • 1
    $\begingroup$ @ZhenLin By a graph I am thinking of two sets $A$ (arrows) and $O$ (objects) and two functions $A\rightarrow O$ (domain and codomain). As can be found in e.g. CWM. $\endgroup$ – H.B. Mar 20 '14 at 14:34
  • 1
    $\begingroup$ @StefanHamcke As you understand now the word 'distinct' was used to say that the arrows are not endomorphisms. $i,j:x\rightarrow y$ is indeed what I mean. $\endgroup$ – H.B. Mar 20 '14 at 14:36
  • 2
    $\begingroup$ @H.B. Then yes, what you have is the category of graphs, almost by definition. $\endgroup$ – Zhen Lin Mar 20 '14 at 14:41

Your Answer

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

Browse other questions tagged or ask your own question.