In the text Joy of Cats, a concrete category over $Set$ is simply a pair $\langle \mathcal C, U \rangle$ consisting of a category $\mathcal C$ and a faithful functor $U\colon \mathcal C \to Set$. But since every small category $\mathcal C$ is isomorphic to a subcategory of $Set$ by $U\colon A\mapsto \{ x\in Ar(\mathcal C)| cod(x)=A \}$ on objects and $U\colon f \mapsto f\circ -$ on arrows, should we say every small category is concrete?

Or should we follow Awodey's statement on p. 14 in his text Category Theory,

A better attempt to capture what is intended by the rather vague notion of a "concrete" category is that arbitrary arrows $f\colon A\to B$ are completely determined by their composites with arrows $x\colon T\to A$ from some "test object" $T$.

  • 1
    $\begingroup$ Awodey's suggestion is not bad, but excludes examples like $\mathbf{Set} \times \mathbf{Set}$. $\endgroup$ – Zhen Lin Mar 21 '14 at 1:11
  • 1
    $\begingroup$ Why not say small categories are concrete? It seems to match with my intuition of concreteness, which is just being modellable in terms of sets. $\endgroup$ – Kevin Carlson Mar 21 '14 at 2:01
  • $\begingroup$ @ZhenLin Wouldn't a "test object" for $\mathbf{Set}\times\mathbf{Set}$ be $\mathbf 2=1\times 1$ since the underlying functor $U\colon\mathbf{Set}\times\mathbf{Set}\to \mathbf{Set}$ is isomorphic to $\mathbf{Set}(\mathbf 2,\square)$, i.e., it is representable. $\endgroup$ – Rachmaninoff Mar 21 '14 at 3:09
  • 3
    $\begingroup$ @Rachmaninoff If I'm not mistaking, in Awodey's approach, one wants that for every $f,g \colon A \to B$, if $fx=gx$ for any $x \colon T \to A$ then $f=g$. So in $\mathbf{Set}\times \mathbf{Set}$, $(1,1)$ cannot distinguish different arrows $(\emptyset,A) \to (\emptyset,A)$ for example (which exist whenever $|A| \geq 2$) because there is no arrow $(1,1) \to (\emptyset,A)$. $\endgroup$ – Pece Mar 21 '14 at 6:25
  • $\begingroup$ @Pece You're right. Thank you for the clarification. $\endgroup$ – Rachmaninoff Mar 21 '14 at 8:26

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.