1
$\begingroup$

I'm getting into a tangle over absolutely elementary definitions in Category theory.

Consider an arbitrary category with some number $n$ of objects (say $n > 2$). Pick two objects in the Category, say $A$ and $B$. Does there (by definition) exist a morphism between $A$ and $B$?

I'd always thought the answer was yes (until I actually started studying some Category theory in earnest - today). I now think the answer is no.

$\endgroup$
  • 3
    $\begingroup$ No. The only morphisms that must exist are identity morphisms. $\endgroup$ – Qiaochu Yuan Sep 29 '17 at 1:08
4
$\begingroup$

Not necessarily. Here's an easy example. Take the category of sets. If $A$ is not empty, then there can be no morphism $A \to \varnothing$.

Other easy (and more severe) examples abound. Take objects to be sets again, but now let $\mathrm{Hom}(A,B)$ be the set of all surjections $A \to B$. Again, this is a category, but it is "lacking" lots of morphisms. Of course, you can play the same game with injective maps.

$\endgroup$
1
$\begingroup$

No, the only morphisms that need to exist are the identity morphisms. So e.g. a two-object category with only identity morphisms (in particular, no morphism from one object to the other) is fine.

$\endgroup$
0
$\begingroup$

You're correct they are not required to exist. I find the easiest example to be an ordered sets like $\mathbb{N}$ with $\mathrm{Hom}(A,B)$ being defined by $\leq$. In this case for $A \neq B$ the existence of $\mathrm{Hom}(A,B)$ strictly forbids the existence of $\mathrm{Hom}(B,A)$.

$\endgroup$
  • $\begingroup$ Note that in that case (if I understand OP's question correctly) there is still a morphism between $A$ and $B$. However, if your order is not a total order then you can find objects with no morphism between them in either direction. $\endgroup$ – Arnaud D. Sep 29 '17 at 8:32
0
$\begingroup$

There are many examples of categories that have this property. Any category with a zero object $z$, such as the category of abelian groups, has this property, because for any two objects $X \to Y$, there always exists the zero map between them, namely the composite $X \to z \to Y$.

Conversely, there are many counterexamples. Any category that has a terminal object $t$ and an initial object $i$ that are not isomorphic, such as the category of sets, then any object $X$ can satisfy at most one of the following two properties:

  • There exists a morphism $t \to X$
  • There exists a morphism $X \to i$.

In particular at least one of $\hom(t, X)$ and $\hom(X, i)$ must be empty.

$\endgroup$

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.