The Stacks project

Definition 7.12.1. Let $\mathcal{C}$ be a category. We say that a family $\{ U_ i \to U\} _{i \in I}$ is an effective epimorphism if all the morphisms $U_ i \to U$ are representable (see Categories, Definition 4.6.4), and for any $X\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the sequence

\[ \xymatrix{ \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(U, X) \ar[r] & \prod \nolimits _{i \in I} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(U_ i, X) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod \nolimits _{(i, j) \in I^2} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(U_ i \times _ U U_ j, X) } \]

is an equalizer diagram. We say that a family $\{ U_ i \to U\} $ is a universal effective epimorphism if for any morphism $V \to U$ the base change $\{ U_ i \times _ U V \to V\} $ is an effective epimorphism.

Comments (5)

Comment #1078 by Rene on

Why does it follow that an effective epimorphism is an epimorphism ?

Comment #1079 by on

Sorry, can you restate the question more clearly? Because if I make a guess for what I think your question means, then it follows immediately from Definitions 7.12.1 and 4.13.1, so you probably have something else in mind.

The terminology (following Artin's notes on Grothendieck topologies) is kind of bad, but I'm afraid we're stuck with it. One small change we could make is to write the definition so it only applies when has fibred products.

Comment #1081 by Rene on

What I dont understand about this definition is how the given equalizer diagram implies epimorphism as given by 003B. Is epimorphism to be assumed as part of the definition ? I find many sources vague on this point. Further there ought to be given a definition of effective epimorphism for simple maps , and not just the covering version.

Comment #1083 by on

OK, first of all, I deduce from your comment that you've understood everything correctly. I think in the case of the definition above the statement is clear. Let us denote the maps in the given family . Namely, the displayed diagram is an equalizer diagram (of sets) if and only if

  1. the first map is injective, in other words, given two maps such that for all , then , and

  2. if for every we are given a morphism such that for every pair the two compositions and are the same, then there is a morphism such that for all .

But, you know, this is just unwinding the definition of equalizer in the category of sets, see Definition 4.10.1. And as the reader can plainly see, the first part is what gives you the epimorphism condition for if the family consists of a single morphism /

Here are the reasons we decided not to have the definition for a single map (and this is what I tried to point out in my previous comment, unsuccessfully):

  1. We already can apply the definition as given to so we don't need to do this (at the very slight cost of having to write the brackets whenever we use the terminology for a single morphism).

  2. If we did this then we might be tempted to call satisfying the current definition a ``family of effective epimorphisms'', but that would be wrong, because it doesn't mean that each is an effective epimorphism.

Moreover, it gets worse when you want to consider also ``universal effective epimorphisms''. I confess I was too stupid to come up with good terminology covering everything unambiguously, so I decided to stick with the above bad terminology which I think is adequate. You can try to find better terminology yourself, but please make sure you cover all cases and for it not to conflict with what is in the literature.

Comment #1084 by on

Hey! I just noticed the products are missing!!! Wow, that sure makes it unreadable! Here is the fix.

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