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I am stuck in trying to interpret a definition in the paper "Categories of continuous functors" by P. Freyd and M. Kelly (click).

They say:

A category $\cal A$ with a proper factorization system $(\mathfrak E, \mathfrak M)$ [i.e. a factorization system where the left class is contained in the class $Epi$ and the right class in the class $Mono$ of monic arrows] has a generator when it has a small full subcategory $\cal G$ such that the family of all morphisms $G\to A$ with domain $G\in\cal G$ is in $\mathfrak E$. If $\cal A$ admits coproducts, then $\cal G$ is a generator iff the canonical arrow $\coprod_{G\to A}G\to A$ lies in $\mathfrak E$ for any $A\in\cal A$.

Edit: Proving that the two conditions are equivalent in presence of coproducts seemed to be easy but trying to reproduce the argument I noticed that the diagram I used wasn't commutative. I wanted to say that the diagram

enter image description here

gives by lifting property the desired arrow to show that each $G\to A$ lies in $\mathfrak E = {}^\perp\mathfrak M$.

What is seems incredible to me is that this notion is the right one to capture the notion of generator, or that of separator, in $\cal A$.

In fact, one of the main point of Freyd-Kelly's paper is that the two notions are not equivalent (as they are stated on the nlab or wikipedia, if I remember well): in a finitely complete -or discrete-cocomplete- category a generator separates arrows; with a particular choice of the factorization system, a separator is a generator.

My problem is that if I interpret "the family of all morphisms $G\to A$ with domain $G\in\cal G$ is in $\mathfrak E$" in the unique possible sense, I can't obtain what I expected: "each arrow $*\to X$ is an epi in $\bf Set$" is a blatantly false statement, even if in that case the terminal object separates arrows.

Can you help me? I feel I'm lost in something easy, but I don't see where.

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If Zhen Lin answered your question to your satisfaction, maybe you could accept his answer? I say this partly because the bots on this site will periodically bring back old questions to the top of the stack if they think they haven't been answered. –  Todd Trimble Jan 28 at 16:26
    
Sure! sorry (to Zhen Lin and to everybody else)! –  tetrapharmakon Jan 28 at 19:09
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1 Answer

up vote 1 down vote accepted

Given an orthogonal (resp. weak) factorisation system $(\mathcal{E}, \mathcal{M})$, we can straightforwardly define when a sink $( U_i \to V : i \in I )$ (where the morphisms $U_i \to V$ may be repeated) is in the left class $\mathcal{E}$: this happens when, for each morphism $X \to Y$ in $\mathcal{M}$, given a commutative square of the form below for each $i$ in $I$, $$\begin{array}{ccc} U_i & \rightarrow & X \\ \downarrow & & \downarrow \\ V & \rightarrow & Y \end{array}$$ there is a unique (resp. at least one) morphism $V \to X$ that makes all the evident diagrams commute simultaneously. In the presence of sufficiently large coproducts, this happens if and only if the amalgamation $\coprod_{i \in I} U_i \to V$ is in $\mathcal{E}$.

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Sure, this is exactly the remark at the end of page 177 in FK paper. But my problem is different: FK say that a generator for a category is the gadget I defined before, and under reasonable hypotheses a generator/generating subcat is the same as a separator/separating subcat. But this is false! The point is a separator in Set, and nevertheless it is not true that any arrow $G=*\to A$ is an epi. On the other hand it is true that the arrow $k_A\colon \coprod_{G\to A}G\to A$ is an epi; where am I lost? –  tetrapharmakon Dec 29 '13 at 14:30
    
Please notice that I edited the OP. Thank you for your time! –  tetrapharmakon Dec 29 '13 at 14:37
    
You have misunderstood something: we are not claiming that each individual morphism is in $\mathcal{E}$, but that they are "jointly" in $\mathcal{E}$. Your "proof" is also mistaken. –  Zhen Lin Dec 29 '13 at 14:39
    
The problem is precisely this, I misunderstood something as I'm left with something which is blatantly false: but how could one interpret the sentence "the family of all morphisms $G\to A$ with domain $G\in\cal G$ is in $\mathfrak E$" in a different manner? The "family" taken as a single set? But it is not an object of $\cal A$, it doesn't make any sense. –  tetrapharmakon Dec 29 '13 at 14:45
    
Ah! I understood. FK define a precise notion of a family "being in $\mathfrak E$", which is precisely the one you are stating. Sorry, this question was boring and ill-posed. –  tetrapharmakon Dec 29 '13 at 14:51
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