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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?

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Your equation is incorrect. You should be looking for $f$ such that $\mathrm{id}_C \circ f = x$. – Zhen Lin Oct 7 '13 at 22:07
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. – Almeo Maus Oct 7 '13 at 22:18
No, $f^*$ is a morphism from $x$ to $\mathrm{id}_C$. – Zhen Lin Oct 7 '13 at 23:00
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. – Almeo Maus Oct 7 '13 at 23:19
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) – Almeo Maus Oct 7 '13 at 23:26
up vote 3 down vote accepted

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).

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MaliceVidrine, you are true! Thank you so much!! – Almeo Maus Oct 8 '13 at 0:06

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