Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For this question I will use $[A, B]$ to represent the class of all morphisms $A\to B$. Also, if $f \in [A, B]$ and $g \in [B, C]$, I will use $f;g$, instead of $g \circ f$, to represent the corresponding element of $[A, C]$.

Now, consider a category $\mathcal{C}$ with the property that in every $[A, B]$ there is a distinguished element $e_{AB}$ with the property that, for any other morphism $f$ with $\mathrm{dom}(f) = B$, the morphism $e_{AB};f$ is also a distinguished morphism. I.e. if $C = \mathrm{dom}(f)$, then $e_{AB};f = e_{AC}$.

My question is, does it follow from the above that, for any object $C$ and any $g \in [C, A]$, we also have that $g;e_{AB} = e_{CB}$?

(In case it matters, you may assume that every $[A, B]$ is a set.)

Thanks!

PS: I have omitted the genesis of this question because it is somewhat convoluted, and I don't think it adds anything to the question, but if anyone is interested, or in case I am wrong about how much it adds to the question, here it is.

I am trying to resolve an ambiguous passage (or at least ambiguous to me) in a book I'm reading. In this passage, the author considers a category $\mathcal{A}$ where each $[A, B]$ (which he assumes is always a set) "has a [unique] distinguished element $e_{AB}$ with the property that the composition of a distinguished morphism with any other morphism is again a distinguished morphism."

I can think of two interpretations for the expression "the composition of a distinguished morphism with any other morphism". The first one of these interprets the word "composition" to subsume both "pre-composition" and "post-composition." In other words, it takes the author's statement to consist of the following two assertions:

1. $\forall f \in [B, C]$, $e_{AB};f = e_{AC}$.
2. $\forall g \in [C, A]$, $g;e_{AB} = e_{CB}$.

The second interpretation is the one that takes the author's statement as being only one of the above, say (1) (duality makes the choice immaterial).

Absent any indication to the contrary, my natural inclination would be to adopt this second interpretation, but it turns out that the author subsequently offers, without proof, several other assertions about $\mathcal{A}$, some of which I can prove (in fact they are quite obvious) only if I assume that both (1) and (2) are true (the remaining I have not yet proved either way).

Assuming that it is indeed the case that the assertions I did prove do require both (1) and (2), I am left with two possibilities: either the first interpretation is what the author meant (i.e. he was asserting both (1) and (2)), or else (1) implies (2).

Hence my original question: does (1) imply (2)?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Take the category with one element, $\mathbb{N}$, and $[\mathbb{N},\mathbb{N}]$ be the set of all functions. Let $e=e_\mathbb{NN}(n)=0$. The, in general, $f;e = e$, but the only thing we can say about $e; f$ is that it is constant.

Now that's the opposite of what you want, but you can now take the opposite category of this category and get a category with the condition that you want.

share|improve this answer
    
Thanks! I confess that when I first read your reply I was slightly miffed, in part because it's such a simple answer ("why didn't I think of it???"), but also, I later realized, because for some reason I was unconsciously hoping that the answer to my question was "yes" + a proof, rather than "no" + a counterexample... Wishful thinking is eeeevil. –  kjo Oct 21 '11 at 3:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.