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In an arbitrary category, we have that even if $X$ and $Y$ have a product $X \times Y$, the natural projections needn't be epimorphisms.

Two questions:

  1. Are there (preferably simple!) conditions we can place on the category such that all the natural projections are, in fact, epimorphisms?

  2. Without assuming anything about the category, is there a (preferably interesting!) weaker property that the natural projections always possess?

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To answer 1 : Note sure about any condition more general than requiring that the category contain a zero object. Not sure about 2 either. –  Prahlad Vaidyanathan Sep 1 '13 at 16:11
    
(1) is not true for $\mathbf{Set}$, so I would say there are no reasonable conditions. –  Zhen Lin Sep 1 '13 at 16:37
    
@ZhenLin, ah of course. Take $X=0$ and $Y$ non-empty, then $X \times Y \rightarrow Y$ isn't epic. I always forget the $X=0$ case.... –  goblin Sep 1 '13 at 16:43
    
@ZhenLin does $\mathrm{Set}^{\mathrm{Op}}$ have the property of interest? I think so but I may be getting confused about the zero case. –  goblin Sep 1 '13 at 16:44
    
Yes, it is true for $\mathbf{Set}^\mathrm{op}$, or more generally, the opposite of any elementary topos. –  Zhen Lin Sep 1 '13 at 17:05

1 Answer 1

up vote 2 down vote accepted

You might like this theorem:

If $\pi _i:P {\to} A_i$ for $i\in I$ is a product and if $i_0 \in I$ is such that, for each $i \in I$, $hom(A_{i0} ,A_i)$ is not empty then $\pi _{i0}$ is a retraction.

In general the $\pi _i$ form an extremal mono-source

See:

Abstract and concrete categories: the joy of cats. Propositions 10.21 and 10.28

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I do like that theorem! Is there a special name for objects $Y$ such that for all objects $X$ we can find a morphism $f : X \rightarrow Y$? The projections associated with any product of such objects are necessarily retractions, by the theorem you give. –  goblin Sep 1 '13 at 17:41
    
I would call $Y$ weakly terminal. But note that this assumption is too strong, we really only need that the factors map to each other. –  Martin Brandenburg Sep 1 '13 at 20:38

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