Let C be a category and let $F$ be a mapping on C. Assume that $F$ is defined for all objects of C. However, on the arrows of C the mapping $F$ does not always behave like a (covariant) functor. For example, it may be the case that for some arrow $f: x \rightarrow y$ in C, the mapping $F$ returns an arrow $F(f): F(x)\rightarrow F(y)$ in C. For another arrow $g:a \rightarrow b$ in C, the mapping $F$ may return an arrow $F(g): F(b)\rightarrow F(a)$ in ${\bf C}^{op}$. And possibly for other arrows of C, the mapping $F$ may not even be defined. Such a mapping is neither covariant not contravariant and it may also be partial. Is there a special name for such mappings?

  • 1
    $\begingroup$ What's the example you have in mind? $\endgroup$
    – varkor
    Commented Dec 16, 2022 at 19:13
  • $\begingroup$ Take for example a partially ordered set $(P,\sqsubseteq)$, which gives rise to a category. Consider a function $F: P \rightarrow P$. Such functions can be monotone, anti-monotone, or neither. It seems that in category theory we are mainly interested in covariant functors (ie., monotone functions) and contravariant functors (ie., anti-monotone ones). Is there a more general notion than just covariant/contravariant? Or is this just called ``a mapping''? I am new to category theory, so I apologize if my question sounds pedantic. $\endgroup$
    – TetraDex
    Commented Dec 17, 2022 at 14:50
  • $\begingroup$ This simply corresponds to a function between the object-classes of the two categories. $\endgroup$
    – varkor
    Commented Dec 17, 2022 at 14:53
  • $\begingroup$ But what if we are also interested in the effect that such a function has on arrows (and only on objects)? $\endgroup$
    – TetraDex
    Commented Dec 17, 2022 at 14:59
  • 3
    $\begingroup$ But in these cases, the posetal structure is irrelevant. It's difficult to suggest a definition without any motivating examples. $\endgroup$
    – varkor
    Commented Dec 17, 2022 at 15:43


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.