0
$\begingroup$

In Awodey's Category Theory (2nd edition), page 229, I read:

the category of elements $J$ of a representable $yC$ has a terminal object, namely the element $1_C \in Hom_{\mathbf{C}}(C,C)$

However, I don't see how this is possible without assuming that all elements of $J$ are split epis.

Indeed, the quoted passage means that: $\forall D \in \mathbf{C}, \forall x \in Hom_{\mathbf{C}}(D,C)$, there is an (unique) arrow from $x$ to $1_C$. Am I correct?

An arrow from $x \in Hom(D,C)$ to $1_C \in Hom(C,C)$ would be of the form $f^*$ with $f \in Hom(C,D)$. We would have: $f^*(x)=x \circ f$. So, $x \circ f = 1_C$, so that $x$ is a split epi.

Could somebody help me find what is wrong with my reasoning, or show how to correct this passage from Awodey?

$\endgroup$
7
  • $\begingroup$ Your equation is incorrect. You should be looking for $f$ such that $\mathrm{id}_C \circ f = x$. $\endgroup$
    – Zhen Lin
    Oct 7 '13 at 22:07
  • $\begingroup$ Thanks ZhenLin for your comment! But if $\mathrm{id}_C \circ f = x$ then $f^*$ is an arrow from $\mathrm{id}_C$ to $x$, right? Though the definition would require the contrary. $\endgroup$
    – Almeo Maus
    Oct 7 '13 at 22:18
  • $\begingroup$ No, $f^*$ is a morphism from $x$ to $\mathrm{id}_C$. $\endgroup$
    – Zhen Lin
    Oct 7 '13 at 23:00
  • $\begingroup$ So, when I said " An arrow from $x \in Hom(D,C)$ to $1_C \in Hom(C,C)$ would be of the form $f^*$ with $f \in Hom(C,D)$ such that $f^*(x)=x \circ f$ " , it was wrong? If so, then I have found the very point on which I am blocking. $\endgroup$
    – Almeo Maus
    Oct 7 '13 at 23:19
  • $\begingroup$ Also, in your comment, $f$ has $C$ as codomain, so $f \in Hom(D,C)$. In that case we would have $f^* \in Hom(Hom(C,C),Hom(D,C))$, as $Hom(.,C)$ is a contravariant functor. And so that $1_C$ cannot be in the domain of $f^*$. So it seems to to me that it is wrong. (though I'm sure it's because I've still missed something) $\endgroup$
    – Almeo Maus
    Oct 7 '13 at 23:26
4
$\begingroup$

My initial answer was based on the same misconception, which I discovered was a misconception by digging farther back to p.196 in Awodey. An arrow $g:\langle f, Hom(A,C)\rangle \to \langle f', Hom(A',C)\rangle$ doesn't come from a $\mathbf{C}$-arrow $A'\to A$, but actually one $g:A\to A'$ after all, with $Hom(g,C)(f')=f$. The composition rule you describe is that of $(1\downarrow Hom(-,C))$, where the category of elements of $yC$ is rather the dual of that comma category (I think).

$\endgroup$
1
  • $\begingroup$ MaliceVidrine, you are true! Thank you so much!! $\endgroup$
    – Almeo Maus
    Oct 8 '13 at 0:06

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.