# Same arrow between distinct objects of a category

I'm a bit bothered by something I've come across, and I'd like to know if the misunderstanding is my own (likely) or the author's (unlikely).

Let $\mathcal{C}$ be a category. An arrow $e : A \to A$ in $\mathcal{C}$ is idempotent if $e \circ e=e$. If $\mathcal{E}$ is a class of (not necessarily all) idempotents in $\mathcal{C}$ then we can form a category $\mathcal{C}[\check{\mathcal{E}}]$ whose objects are the members of $\mathcal{E}$ and whose arrows $f : (e:A \to A) \to (d:B \to B)$ are arrows $f : A \to B$ in $\mathcal{C}$ satisfying $d \circ f \circ e=f$.

Something confuses me here. Say $1_A, 1_B \in \mathcal{E}$ and $(e,d) \ne (1_A,1_B)$. We have $d\circ f \circ e=f=1_B \circ f \circ 1_A$. So in the definition of $\mathcal{C}[\check{\mathcal{E}}]$ given above, $f$ is both an arrow $e \to d$ and an arrow $1_A \to 1_B$.

To me this seems absurd. But is it? Can we declare $f : e \to d$ and $f : 1_A \to 1_B$ to be distinct arrows in $\mathcal{C}[\check{\mathcal{E}}]$ despite having the same underlying arrow in $\mathcal{C}$?

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Sure. We just stick extra labels onto each arrow indicating its new domain and codomain. (A morphism in $\mathcal{C}[\check{\mathcal{E}}]$ is formally a triplet $(f, e, d)$ where etc.) –  Zhen Lin Aug 7 '12 at 15:57
@ZhenLin: Yup, thanks. I had a momentary lapse of common sense but it's been resolved now :) Thanks. –  Clive Newstead Aug 7 '12 at 15:58

For a motivating example, consider for example category $Set$, and the objects $\{0\}$, $\{0,1\}$, and the arrows $f,g$ from $\{0\}$ to $\{0\}$ and to $\{0,1\}$, respectively, both of which correspond to the map $x\mapsto 0$. They both have the same "underlying mechanism", interpretation, but are nonetheless distinct arrows.
I thought of this but for some reason it didn't click because I thought $f,g$ having different codomains made it different... but, of course, so do the ones in my question. I think I'm close to being convinced, so I'll accept your answer when it lets me $-$ thanks! –  Clive Newstead Aug 7 '12 at 15:56
There is a bit of "abuse of notation" going on here. To begin with $f:A \rightarrow B$ uses $f$ to refer to an arrow in category $\mathcal{C}$. We are then reusing $f$ to label arrows in the new category $\mathcal{C}[\check{\mathcal{E}}]$. As you point out this does not give unique labels for the arrows of the new category, but it would be cumbersome to continually adjoin the domain and codomain to the labelling, just for the sake of making unique names.